question
string
answer
string
context
string
dialect
string
spider_hardness
string
extended_hardness
string
classification_status
string
syntax_status
string
ddl_syntax_status
string
schema_status
string
syntax_errors
list
ddl_syntax_errors
list
schema_errors
list
classification_errors
list
status_history
list
correction_status
string
component1
int64
component2
int64
others
int64
validation_source
dict
Find the title of course whose prerequisite is course Differential Geometry.
SELECT title FROM course WHERE course_id IN (SELECT T1.course_id FROM prereq AS T1 JOIN course AS T2 ON T1.prereq_id = T2.course_id WHERE T2.title = 'Differential Geometry')
CREATE TABLE prereq (course_id VARCHAR(255), prereq_id VARCHAR(255)); CREATE TABLE course (title VARCHAR(255), course_id VARCHAR(255))
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
1
1
0
{ "split": "train", "row_index": 817 }
What is the title of the course that was offered at building Chandler during the fall semester in the year of 2010?
SELECT T1.title FROM course AS T1 JOIN SECTION AS T2 ON T1.course_id = T2.course_id WHERE `building` = 'Chandler' AND `semester` = 'Fall' AND `YEAR` = 2010
CREATE TABLE course (title VARCHAR(255), course_id VARCHAR(255)); CREATE TABLE SECTION (course_id VARCHAR(255), `building` VARCHAR(255), `semester` VARCHAR(255), `YEAR` INTEGER)
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 819 }
What is the course title of the prerequisite of course Mobile Computing?
SELECT title FROM course WHERE course_id IN (SELECT T1.prereq_id FROM prereq AS T1 JOIN course AS T2 ON T1.course_id = T2.course_id WHERE T2.title = 'Mobile Computing')
CREATE TABLE course (title VARCHAR(255), course_id VARCHAR(255)); CREATE TABLE prereq (prereq_id VARCHAR(255), course_id VARCHAR(255))
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
1
1
0
{ "split": "train", "row_index": 823 }
Find the names of all instructors who have taught some course and the course_id.
SELECT T1.name, T2.course_id FROM instructor AS T1 JOIN teaches AS T2 ON T1.ID = T2.ID
CREATE TABLE instructor (ID VARCHAR(255), name VARCHAR(255)); CREATE TABLE teaches (ID VARCHAR(255), course_id INTEGER)
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
1
0
1
{ "split": "train", "row_index": 839 }
Find the names of all instructors in the Art department who have taught some course and the course_id.
SELECT T1.name, T2.course_id FROM instructor AS T1 JOIN teaches AS T2 ON T1.ID = T2.ID WHERE T1.dept_name = 'Art'
CREATE TABLE instructor (ID VARCHAR(255), dept_name VARCHAR(255), name VARCHAR(255)); CREATE TABLE teaches (ID VARCHAR(255), course_id INTEGER)
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 840 }
Find all the policy type codes associated with the customer "Dayana Robel"
SELECT policy_type_code FROM policies AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id WHERE t2.customer_details = 'Dayana Robel'
CREATE TABLE customers (customer_id VARCHAR(255), customer_details VARCHAR(255)); CREATE TABLE policies (customer_id VARCHAR(255), policy_type_code VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
0
{ "split": "train", "row_index": 867 }
Find the total amount claimed in the most recently created document.
SELECT SUM(t1.amount_claimed) FROM claim_headers AS t1 JOIN claims_documents AS t2 ON t1.claim_header_id = t2.claim_id WHERE t2.created_date = (SELECT created_date FROM claims_documents ORDER BY created_date LIMIT 1)
CREATE TABLE claim_headers (amount_claimed INTEGER, claim_header_id VARCHAR(255)); CREATE TABLE claims_documents (claim_id VARCHAR(255), created_date VARCHAR(255))
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
1
0
{ "split": "train", "row_index": 871 }
What is the name of the customer who has made the largest amount of claim in a single claim?
SELECT t3.customer_details FROM claim_headers AS t1 JOIN policies AS t2 ON t1.policy_id = t2.policy_id JOIN customers AS t3 ON t2.customer_id = t3.customer_id WHERE t1.amount_claimed = (SELECT MAX(amount_claimed) FROM claim_headers)
CREATE TABLE customers (customer_details VARCHAR(255), customer_id VARCHAR(255)); CREATE TABLE policies (policy_id VARCHAR(255), customer_id VARCHAR(255)); CREATE TABLE claim_headers (policy_id VARCHAR(255), amount_claimed INTEGER);
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
3
1
0
{ "split": "train", "row_index": 872 }
Find the names of customers who have no policies associated.
SELECT customer_details FROM customers EXCEPT SELECT t2.customer_details FROM policies AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id
CREATE TABLE customers (customer_details VARCHAR(255), customer_id VARCHAR(255)); CREATE TABLE policies (customer_id VARCHAR(255), policy_id INTEGER PRIMARY KEY AUTO_INCREMENT)
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 874 }
Which customer is associated with the latest policy?
SELECT t2.customer_details FROM policies AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id WHERE t1.start_date = (SELECT MAX(start_date) FROM policies)
CREATE TABLE customers (customer_details VARCHAR(255), customer_id VARCHAR(255)); CREATE TABLE policies (customer_id VARCHAR(255), start_date INTEGER);
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
1
0
{ "split": "train", "row_index": 885 }
Show all product names without an order.
SELECT product_name FROM Products EXCEPT SELECT T1.product_name FROM Products AS T1 JOIN Order_items AS T2 ON T1.product_id = T2.product_id
CREATE TABLE Products (product_name VARCHAR(255), product_id VARCHAR(255)); CREATE TABLE Order_items (product_id VARCHAR(255), order_item_id INTEGER PRIMARY KEY AUTO_INCREMENT)
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 915 }
What is the theme and artist name for the exhibition with a ticket price higher than the average?
SELECT T1.theme, T2.name FROM exhibition AS T1 JOIN artist AS T2 ON T1.artist_id = T2.artist_id WHERE T1.ticket_price > (SELECT AVG(ticket_price) FROM exhibition)
CREATE TABLE exhibition (theme VARCHAR(255), artist_id VARCHAR(255), ticket_price INTEGER); CREATE TABLE artist (name VARCHAR(255), artist_id VARCHAR(255))
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
1
1
{ "split": "train", "row_index": 955 }
Find each target user's name and average trust score.
SELECT T1.name, AVG(trust) FROM useracct AS T1 JOIN trust AS T2 ON T1.u_id = T2.target_u_id GROUP BY T2.target_u_id
CREATE TABLE trust (target_u_id VARCHAR(255), trust INTEGER); CREATE TABLE useracct (name VARCHAR(255), u_id VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 987 }
Find the name of the target user with the lowest trust score.
SELECT T1.name FROM useracct AS T1 JOIN trust AS T2 ON T1.u_id = T2.target_u_id ORDER BY T2.trust_score LIMIT 1
CREATE TABLE trust (target_u_id VARCHAR(255), trust_score INTEGER); CREATE TABLE useracct (name VARCHAR(255), u_id VARCHAR(255))
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
3
0
0
{ "split": "train", "row_index": 988 }
List the distinct hometowns that are not associated with any gymnast.
SELECT DISTINCT Hometown FROM people EXCEPT SELECT DISTINCT T2.Hometown FROM gymnast AS T1 JOIN people AS T2 ON T1.Gymnast_ID = T2.People_ID
CREATE TABLE gymnast (Gymnast_ID VARCHAR(255), name VARCHAR(255)); CREATE TABLE people (Hometown VARCHAR(255), People_ID VARCHAR(255))
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 1024 }
Show all member names who are not in charge of any event.
SELECT member_name FROM member EXCEPT SELECT T1.member_name FROM member AS T1 JOIN party_events AS T2 ON T1.member_id = T2.member_in_charge_id
CREATE TABLE member (member_name VARCHAR(255), member_id VARCHAR(255)); CREATE TABLE party_events (member_in_charge_id VARCHAR(255), event_id VARCHAR(255))
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 1230 }
What are the ids and details of events that have more than one participants?
SELECT T1.event_id, T1.event_details FROM EVENTS AS T1 JOIN Participants_in_Events AS T2 ON T1.event_id = T2.Event_ID GROUP BY T1.event_id HAVING COUNT(*) > 1
CREATE TABLE EVENTS (event_id VARCHAR(255), event_details VARCHAR(255)); CREATE TABLE Participants_in_Events (Event_ID VARCHAR(255), participant_id VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 1291 }
How many drivers participated in the race Australian Grand Prix held in 2009?
SELECT COUNT(*) FROM results AS T1 JOIN races AS T2 ON T1.raceid = T2.raceid WHERE T2.name = 'Australian Grand Prix' AND `year` = 2009
CREATE TABLE races (raceid VARCHAR(255), name VARCHAR(255), `year` INTEGER); CREATE TABLE results (raceid VARCHAR(255), driver_id VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 1314 }
How many degrees were conferred in "San Jose State University" in 2000?
SELECT T2.degrees FROM campuses AS T1 JOIN degrees AS T2 ON T1.Id = T2.campus WHERE T1.campus = 'San Jose State University' AND T2.year = 2000
CREATE TABLE degrees (Id VARCHAR(255), campus VARCHAR(255), year INTEGER, degrees INTEGER); CREATE TABLE campuses (Id VARCHAR(255), campus VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 1421 }
How many faculty lines are there in the university that conferred the most number of degrees in year 2002?
SELECT T2.faculty FROM campuses AS T1 JOIN faculty AS T2 ON T1.id = T2.campus AND T2.year = 2002 JOIN degrees AS T3 ON T1.id = T3.campus AND T3.year = 2002 WHERE T2.year = 2002 ORDER BY T3.degrees DESC LIMIT 1
CREATE TABLE campuses (id VARCHAR(255), name VARCHAR(255)); CREATE TABLE faculty (faculty VARCHAR(255), campus VARCHAR(255), year INTEGER); CREATE TABLE degrees (Id VARCHAR(255), campus VARCHAR(255), year INTEGER, degrees INTEGER)
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
5
0
0
{ "split": "train", "row_index": 1427 }
How many faculty lines are there in the university that conferred the least number of degrees in year 2001?
SELECT T2.faculty FROM campuses AS T1 JOIN faculty AS T2 ON T1.id = T2.campus AND T2.year = T3.year JOIN degrees AS T3 ON T1.id = T3.campus AND T2.year = T3.year WHERE T2.year = 2001 ORDER BY T3.degrees LIMIT 1
CREATE TABLE campuses (id VARCHAR(255), name VARCHAR(255)); CREATE TABLE faculty (faculty VARCHAR(255), campus VARCHAR(255), year INTEGER); CREATE TABLE degrees (id VARCHAR(255), campus VARCHAR(255), year INTEGER, degrees INTEGER)
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
5
0
0
{ "split": "train", "row_index": 1428 }
Find the name of the candidates whose oppose percentage is the lowest for each sex.
SELECT t1.name, t1.sex, MIN(t2.oppose_rate) FROM people AS t1 JOIN candidate AS t2 ON t1.people_id = t2.people_id GROUP BY t1.sex
CREATE TABLE candidate (people_id VARCHAR(255), oppose_rate INTEGER); CREATE TABLE people (name VARCHAR(255), sex VARCHAR(255), people_id VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 1445 }
Find the names of all reviewers who have ratings with a NULL value for the date.
SELECT DISTINCT `name` FROM `Reviewer` AS T1 JOIN `Rating` AS T2 ON T1.`rID` = T2.`rID` WHERE `ratingDate` IS NULL
CREATE TABLE `Rating` (`rID` VARCHAR(255), `ratingDate` DATE); CREATE TABLE `Reviewer` (`rID` VARCHAR(255), `name` VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
0
{ "split": "train", "row_index": 1467 }
What is the average rating stars and title for the oldest movie?
SELECT AVG(T1.stars), T2.title FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID = T2.mID WHERE T2.year = (SELECT MIN(year) FROM Movie)
CREATE TABLE Movie (title VARCHAR(255), mID VARCHAR(255), year INTEGER); CREATE TABLE Rating (stars INTEGER, mID VARCHAR(255))
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
1
1
{ "split": "train", "row_index": 1468 }
Show all storm names except for those with at least two affected regions.
SELECT name FROM storm EXCEPT SELECT T1.name FROM storm AS T1 JOIN affected_region AS T2 ON T1.storm_id = T2.storm_id GROUP BY T1.storm_id HAVING COUNT(*) >= 2
CREATE TABLE storm (name VARCHAR(255), storm_id VARCHAR(255)); CREATE TABLE affected_region (storm_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 1603 }
How long does student Linda Smith spend on the restaurant in total?
SELECT SUM(Visits_Restaurant.Spent) FROM Student JOIN Visits_Restaurant ON Student.StuID = Visits_Restaurant.StuID WHERE Student.Fname = 'Linda' AND Student.Lname = 'Smith'
CREATE TABLE Visits_Restaurant (Spent INTEGER, StuID INTEGER); CREATE TABLE Student (StuID INTEGER, Fname VARCHAR(255), Lname VARCHAR(255), Spent INTEGER)
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 1678 }
At which restaurant did the students spend the least amount of time? List restaurant and the time students spent on in total.
SELECT Restaurant.ResName, SUM(Visits_Restaurant.Spent) FROM Visits_Restaurant JOIN Restaurant ON Visits_Restaurant.ResID = Restaurant.ResID GROUP BY Restaurant.ResID ORDER BY SUM(Visits_Restaurant.Spent) LIMIT 1
CREATE TABLE Visits_Restaurant (Id VARCHAR(255), ResID VARCHAR(255), Spent INTEGER); CREATE TABLE Restaurant (Id VARCHAR(255), ResID VARCHAR(255), ResName VARCHAR(255))
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
4
0
2
{ "split": "train", "row_index": 1681 }
Retrieve the title of the paper that has the largest number of authors.
SELECT t2.title FROM authorship AS t1 JOIN papers AS t2 ON t1.paperid = t2.paperid WHERE t1.authorder = (SELECT MAX(authorder) FROM authorship)
CREATE TABLE authorship (paperid VARCHAR(255), authorder INTEGER); CREATE TABLE papers (title VARCHAR(255), paperid VARCHAR(255))
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
1
0
{ "split": "train", "row_index": 1726 }
Which film is rented at a fee of 0.99 and has less than 3 in the inventory? List the film title and id.
SELECT f.title, f.film_id FROM film AS f WHERE f.rental_rate = '0.99' AND (SELECT COUNT(*) FROM inventory i WHERE i.film_id = f.film_id) < 3;
CREATE TABLE film (title VARCHAR(255), film_id VARCHAR(255), rental_rate VARCHAR(255)); CREATE TABLE inventory (film_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT);
mysql
null
extreme
extended
valid
valid
valid
[]
[]
[]
[ "Unsupported column expression: (SELECT COUNT(*) FROM inventory AS i WHERE i.film_id = f.film_id)" ]
[ "validation_failed", "correction_corrected", "extended", "extended" ]
corrected
null
null
null
{ "split": "train", "row_index": 1739 }
Find all the payment dates for the payments with an amount larger than 10 and the payments handled by a staff person with the first name Elsa.
SELECT payment_date FROM payment WHERE amount > 10 UNION SELECT T1.payment_date FROM payment AS T1 JOIN staff AS T2 ON T1.staff_id = T2.staff_id WHERE T2.first_name = 'Elsa'
CREATE TABLE payment (payment_date VARCHAR(255), staff_id VARCHAR(255), amount INTEGER); CREATE TABLE staff (staff_id VARCHAR(255), first_name VARCHAR(255));
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
1
1
0
{ "split": "train", "row_index": 1742 }
Which customer, who has made at least one payment, has spent the least money? List his or her first name, last name, and the id.
SELECT T1.first_name, T1.last_name, T1.customer_id FROM customer AS T1 JOIN payment AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY SUM(T2.amount) LIMIT 1
CREATE TABLE payment (customer_id VARCHAR(255), amount INTEGER); CREATE TABLE customer (first_name VARCHAR(255), last_name VARCHAR(255), customer_id VARCHAR(255));
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
4
0
1
{ "split": "train", "row_index": 1750 }
Where is store 1 located?
SELECT T2.address FROM store AS T1 JOIN address AS T2 ON T1.address_id = T2.address_id WHERE store_id = 1
CREATE TABLE store (address_id VARCHAR(255), store_id INTEGER); CREATE TABLE address (address VARCHAR(255), address_id VARCHAR(255));
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
0
{ "split": "train", "row_index": 1757 }
Find the name of customers who do not have a loan with a type of Mortgages.
SELECT cust_name FROM customer EXCEPT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE T2.loan_type = 'Mortgages'
CREATE TABLE loan (cust_id VARCHAR(255), loan_type VARCHAR(255)); CREATE TABLE customer (cust_name VARCHAR(255), cust_id VARCHAR(255));
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 1790 }
Find the name of customers who have loans of both Mortgages and Auto.
SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE T2.loan_type = 'Mortgages' INTERSECT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE T2.loan_type = 'Auto'
CREATE TABLE customer (cust_name VARCHAR(255), cust_id VARCHAR(255)); CREATE TABLE loan (cust_id VARCHAR(255), loan_type VARCHAR(255));
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_failed", "correction_corrected", "extended", "classified" ]
corrected
2
1
0
{ "split": "train", "row_index": 1791 }
Find the total amount of loans offered by each bank branch.
SELECT SUM(T2.amount) AS total_amount, T1.bname FROM bank AS T1 JOIN loan AS T2 ON T1.branch_id = T2.branch_id GROUP BY T1.bname
CREATE TABLE loan (branch_id VARCHAR(255), amount INTEGER); CREATE TABLE bank (bname VARCHAR(255), branch_id VARCHAR(255));
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 1800 }
Find the the name of the customers who have a loan with amount more than 3000.
SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE T2.amount > 3000
CREATE TABLE customer (cust_name VARCHAR(255), cust_id VARCHAR(255)); CREATE TABLE loan (cust_id VARCHAR(255), amount INTEGER);
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
0
{ "split": "train", "row_index": 1807 }
What are the id and first name of the student whose addresses have the highest average monthly rental?
SELECT T1.student_id, T2.first_name FROM Student_Addresses AS T1 JOIN Students AS T2 ON T1.student_id = T2.student_id GROUP BY T1.student_id ORDER BY AVG(monthly_rental) DESC LIMIT 1
CREATE TABLE Students (first_name VARCHAR(255), student_id VARCHAR(255)); CREATE TABLE Student_Addresses (student_id VARCHAR(255), monthly_rental INTEGER);
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
4
0
1
{ "split": "train", "row_index": 1835 }
Find the id and city of the student address with the highest average monthly rental.
SELECT T2.address_id, T1.city FROM Addresses AS T1 JOIN Student_Addresses AS T2 ON T1.address_id = T2.address_id GROUP BY T2.address_id ORDER BY AVG(monthly_rental) DESC LIMIT 1
CREATE TABLE Student_Addresses (address_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE Addresses (city VARCHAR(255), address_id VARCHAR(255), monthly_rental INTEGER);
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
4
0
1
{ "split": "train", "row_index": 1836 }
Find the start and end dates of behavior incidents of students with last name "Fahey".
SELECT T1.date_incident_start, T1.date_incident_end FROM Behavior_Incident AS T1 JOIN Students AS T2 ON T1.student_id = T2.student_id WHERE T2.last_name = 'Fahey'
CREATE TABLE Behavior_Incident (date_incident_start VARCHAR(255), date_incident_end VARCHAR(255), student_id VARCHAR(255)); CREATE TABLE Students (student_id VARCHAR(255), last_name VARCHAR(255));
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 1841 }
Find the start and end dates of detentions of teachers with last name "Schultz".
SELECT T1.datetime_detention_start, T1.datetime_detention_end FROM Detention AS T1 JOIN Teachers AS T2 ON T1.teacher_id = T2.teacher_id WHERE T2.last_name = 'Schultz'
CREATE TABLE Detention (datetime_detention_start VARCHAR(255), datetime_detention_end VARCHAR(255), teacher_id VARCHAR(255)); CREATE TABLE Teachers (teacher_id VARCHAR(255), last_name VARCHAR(255));
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 1842 }
What are the id and zip code of the address with the highest monthly rental?
SELECT T2.address_id, T1.zip_postcode FROM Addresses AS T1 JOIN Student_Addresses AS T2 ON T1.address_id = T2.address_id ORDER BY monthly_rental DESC LIMIT 1
CREATE TABLE Student_Addresses (address_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE Addresses (zip_postcode VARCHAR(255), address_id VARCHAR(255), monthly_rental INTEGER);
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
3
0
1
{ "split": "train", "row_index": 1843 }
Find the last names of teachers who are not involved in any detention.
SELECT last_name FROM Teachers EXCEPT SELECT T1.last_name FROM Teachers AS T1 JOIN Detention AS T2 ON T1.teacher_id = T2.teacher_id
CREATE TABLE Teachers (last_name VARCHAR(255), teacher_id VARCHAR(255)); CREATE TABLE Detention (teacher_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT);
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 1851 }
Find the id and local authority of the station with has the highest average high temperature.
SELECT t2.id, t2.local_authority FROM weekly_weather AS t1 JOIN station AS t2 ON t1.station_id = t2.id GROUP BY t1.station_id ORDER BY AVG(high_temperature) DESC LIMIT 1
CREATE TABLE weekly_weather (station_id VARCHAR(255), high_temperature INTEGER); CREATE TABLE station (id VARCHAR(255), local_authority VARCHAR(255));
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
4
0
1
{ "split": "train", "row_index": 1896 }
Find the number of professors in accounting department.
SELECT COUNT(*) FROM professor AS T1 JOIN department AS T2 ON T1.dept_code = T2.dept_code WHERE `DEPT_NAME` = 'Accounting'
CREATE TABLE professor (dept_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE department (dept_code VARCHAR(255), `DEPT_NAME` VARCHAR(255));
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
0
{ "split": "train", "row_index": 1900 }
Find the number of classes in each department.
SELECT COUNT(*), dept_code FROM CLASS AS T1 JOIN course AS T2 ON T1.crs_code = T2.crs_code GROUP BY dept_code
CREATE TABLE CLASS (crs_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE course (crs_code VARCHAR(255), dept_code VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 1916 }
How many courses are offered by the Computer Info. Systems department?
SELECT COUNT(*) FROM department AS T1 JOIN course AS T2 ON T1.dept_code = T2.dept_code WHERE dept_name = 'Computer Info. Systems'
CREATE TABLE course (dept_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE department (dept_code VARCHAR(255), dept_name VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
0
{ "split": "train", "row_index": 1930 }
What are the first names of the professors who do not teach a class.
SELECT emp_fname FROM employee WHERE emp_jobcode = 'PROF' EXCEPT SELECT T1.emp_fname FROM employee AS T1 JOIN CLASS AS T2 ON T1.emp_num = T2.prof_num
CREATE TABLE employee (emp_fname VARCHAR(255), emp_jobcode VARCHAR(255), emp_num VARCHAR(255)); CREATE TABLE CLASS (prof_num VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
1
1
0
{ "split": "train", "row_index": 1950 }
What is the department name of the students with lowest gpa belongs to?
SELECT T2.dept_name FROM student AS T1 JOIN department AS T2 ON T1.dept_code = T2.dept_code ORDER BY stu_gpa LIMIT 1
CREATE TABLE student (dept_code VARCHAR(255), stu_gpa INTEGER); CREATE TABLE department (dept_name VARCHAR(255), dept_code VARCHAR(255))
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
3
0
0
{ "split": "train", "row_index": 1963 }
Find the names of furnitures whose prices are lower than the highest price.
SELECT t1.name FROM furniture AS t1 JOIN furniture_manufacte AS t2 ON t1.Furniture_ID = t2.Furniture_ID WHERE t2.Price_in_Dollar < (SELECT MAX(Price_in_Dollar) FROM furniture_manufacte)
CREATE TABLE furniture_manufacte (Furniture_ID VARCHAR(255), Price_in_Dollar INTEGER); CREATE TABLE furniture (name VARCHAR(255), Furniture_ID VARCHAR(255))
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
1
0
{ "split": "train", "row_index": 2013 }
display the department name and number of employees in each of the department.
SELECT T2.department_name, COUNT(*) FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id GROUP BY T2.department_name
CREATE TABLE employees (department_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE departments (department_id VARCHAR(255), department_name VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 2081 }
Find the names of programs that are never broadcasted in the morning.
SELECT name FROM program EXCEPT SELECT t1.name FROM program AS t1 JOIN broadcast AS t2 ON t1.program_id = t2.program_id WHERE t2.Time_of_day = 'Morning'
CREATE TABLE broadcast (program_id VARCHAR(255), Time_of_day VARCHAR(255)); CREATE TABLE program (name VARCHAR(255), program_id VARCHAR(255));
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 2203 }
Find the personal names of students not enrolled in any course.
SELECT personal_name FROM Students EXCEPT SELECT T1.personal_name FROM Students AS T1 JOIN Student_Course_Enrolment AS T2 ON T1.student_id = T2.student_id
CREATE TABLE Student_Course_Enrolment (student_id VARCHAR(255), enrollment_date DATE); CREATE TABLE Students (personal_name VARCHAR(255), student_id VARCHAR(255));
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 2248 }
Which claims caused more than 2 settlements or have the maximum claim value? List the date the claim was made and the claim id.
SELECT T1.Date_Claim_Made, T1.Claim_id FROM Claims AS T1 JOIN Settlements AS T2 ON T1.Claim_id = T2.Claim_id GROUP BY T1.Claim_id HAVING COUNT(*) > 2 UNION SELECT T1.Date_Claim_Made, T1.Claim_id FROM Claims AS T1 JOIN Settlements AS T2 ON T1.Claim_id = T2.Claim_id WHERE T1.Amount_Claimed = (SELECT MAX(Amount_Claimed) F...
CREATE TABLE Settlements (Claim_id VARCHAR(255), amount INTEGER); CREATE TABLE Claims (Date_Claim_Made VARCHAR(255), Claim_id VARCHAR(255), Amount_Claimed INTEGER);
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
1
1
{ "split": "train", "row_index": 2252 }
Which customer had at least 2 policies but did not file any claims? List the customer details and id.
SELECT T1.customer_details, T1.customer_id FROM Customers AS T1 JOIN Customer_Policies AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id HAVING COUNT(*) >= 2 EXCEPT SELECT T1.customer_details, T1.customer_id FROM Customers AS T1 JOIN Customer_Policies AS T2 ON T1.customer_id = T2.customer_id JOIN Claims ...
CREATE TABLE Customers (customer_details VARCHAR(255), customer_id VARCHAR(255)); CREATE TABLE Customer_Policies (customer_id VARCHAR(255), policy_id VARCHAR(255)); CREATE TABLE Claims (policy_id VARCHAR(255), amount INTEGER);
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
1
1
{ "split": "train", "row_index": 2253 }
How many settlements does each claim correspond to? List the claim id and the number of settlements.
SELECT T1.`Claim_id`, COUNT(*) FROM Claims AS T1 JOIN Settlements AS T2 ON T1.`Claim_id` = T2.claim_id GROUP BY T1.`Claim_id`
CREATE TABLE Settlements (claim_id VARCHAR(255), settlement_date DATE); CREATE TABLE Claims (`Claim_id` VARCHAR(255), amount INTEGER);
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 2259 }
Who are the customers that had more than 1 policy? List the customer details and id.
SELECT T1.customer_details, T1.customer_id FROM Customers AS T1 JOIN Customer_Policies AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id HAVING COUNT(*) > 1
CREATE TABLE Customers (customer_details VARCHAR(255), customer_id VARCHAR(255)); CREATE TABLE Customer_Policies (customer_id VARCHAR(255), policy_id VARCHAR(255));
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 2264 }
List the details of the customers who do not have any policies.
SELECT customer_details FROM Customers EXCEPT SELECT T1.customer_details FROM Customers AS T1 JOIN Customer_Policies AS T2 ON T1.customer_id = T2.customer_id
CREATE TABLE Customer_Policies (customer_id VARCHAR(255), policy_id VARCHAR(255)); CREATE TABLE Customers (customer_details VARCHAR(255), customer_id VARCHAR(255));
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 2273 }
List the date the claim was made, the date it was settled and the amount settled for all the claims which had exactly one settlement.
SELECT T1.claim_id, T1.date_claim_made, T1.Date_Claim_Settled FROM Claims AS T1 JOIN Settlements AS T2 ON T1.claim_id = T2.Claim_id GROUP BY T1.claim_id HAVING COUNT(*) = 1
CREATE TABLE Claims (claim_id VARCHAR(255), date_claim_made VARCHAR(255), Date_Claim_Settled VARCHAR(255)); CREATE TABLE Settlements (Claim_id VARCHAR(255), settlement_amount INTEGER);
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 2274 }
what is the name and position of the head whose department has least number of employees?
SELECT T2.name, T2.position FROM department AS T1 JOIN physician AS T2 ON T1.head = T2.EmployeeID GROUP BY `departmentID` ORDER BY COUNT(`departmentID`) LIMIT 1
CREATE TABLE department (head VARCHAR(255), `departmentID` INTEGER); CREATE TABLE physician (name VARCHAR(255), position VARCHAR(255), EmployeeID VARCHAR(255));
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
4
0
1
{ "split": "train", "row_index": 2278 }
What are names of patients who made an appointment?
SELECT T2.name FROM appointment AS T1 JOIN patient AS T2 ON T1.patient = T2.ssn
CREATE TABLE appointment (patient VARCHAR(255), appointment_date DATE); CREATE TABLE patient (ssn VARCHAR(255), name VARCHAR(255));
mysql
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
1
0
0
{ "split": "train", "row_index": 2279 }
what are name and phone number of patients who had more than one appointment?
SELECT T2.name, T2.phone FROM appointment AS T1 JOIN patient AS T2 ON T1.patient = T2.ssn GROUP BY T1.patient HAVING COUNT(*) > 1
CREATE TABLE appointment (patient VARCHAR(255), appointment_date DATE); CREATE TABLE patient (ssn VARCHAR(255), name VARCHAR(255), phone VARCHAR(255));
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 2280 }
List the name of physicians who never took any appointment.
SELECT name FROM physician EXCEPT SELECT T2.name FROM appointment AS T1 JOIN physician AS T2 ON T1.Physician = T2.EmployeeID
CREATE TABLE physician (name VARCHAR(255), EmployeeID VARCHAR(255)); CREATE TABLE appointment (Physician VARCHAR(255), appointment_date DATE);
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 2283 }
Find the name of medication used on the patient who stays in room 111?
SELECT T4.name FROM stay AS T1 JOIN patient AS T2 ON T1.Patient = T2.SSN JOIN Prescribes AS T3 ON T3.Patient = T2.SSN JOIN Medication AS T4 ON T3.Medication = T4.Code WHERE `room` = 111
CREATE TABLE Prescribes (Patient VARCHAR(255), Medication VARCHAR(255)); CREATE TABLE Medication (name VARCHAR(255), Code VARCHAR(255)); CREATE TABLE patient (SSN VARCHAR(255), name VARCHAR(255)); CREATE TABLE stay (Patient VARCHAR(255), `room` INTEGER);
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
4
0
0
{ "split": "train", "row_index": 2288 }
List the names of departments where some physicians are primarily affiliated with.
SELECT DISTINCT T2.name FROM affiliated_with AS T1 JOIN department AS T2 ON T1.department = T2.departmentid WHERE PrimaryAffiliation = 1
CREATE TABLE affiliated_with (department VARCHAR(255), PrimaryAffiliation INTEGER); CREATE TABLE department (name VARCHAR(255), departmentid VARCHAR(255));
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
0
{ "split": "train", "row_index": 2308 }
Find the names of all procedures which cost more than 1000 but which physician John Wen was not trained in?
SELECT name FROM procedures WHERE cost > 1000 EXCEPT SELECT T3.name FROM physician AS T1 JOIN trained_in AS T2 ON T1.employeeid = T2.physician JOIN procedures AS T3 ON T3.code = T2.treatment WHERE T1.name = 'John Wen'
CREATE TABLE trained_in (physician VARCHAR(255), treatment VARCHAR(255)); CREATE TABLE procedures (name VARCHAR(255), cost INTEGER, code VARCHAR(255)); CREATE TABLE physician (employeeid VARCHAR(255), name VARCHAR(255))
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
1
1
0
{ "split": "train", "row_index": 2318 }
Find the names of all procedures such that the cost is less than 5000 and physician John Wen was trained in.
SELECT name FROM procedures WHERE cost < 5000 INTERSECT SELECT T3.name FROM physician AS T1 JOIN trained_in AS T2 ON T1.employeeid = T2.physician JOIN procedures AS T3 ON T3.code = T2.treatment WHERE T1.name = 'John Wen'
CREATE TABLE trained_in (physician VARCHAR(255), treatment VARCHAR(255)); CREATE TABLE procedures (name VARCHAR(255), cost INTEGER, code VARCHAR(255)); CREATE TABLE physician (employeeid VARCHAR(255), name VARCHAR(255))
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
1
1
0
{ "split": "train", "row_index": 2319 }
Find the names of patients who are not using the medication of Procrastin-X.
SELECT name FROM patient EXCEPT SELECT T1.name FROM patient AS T1 JOIN Prescribes AS T2 ON T2.Patient = T1.SSN JOIN Medication AS T3 ON T2.Medication = T3.Code WHERE T3.name = 'Procrastin-X'
CREATE TABLE Prescribes (Patient VARCHAR(255), Medication VARCHAR(255)); CREATE TABLE Medication (Code VARCHAR(255), name VARCHAR(255)); CREATE TABLE patient (name VARCHAR(255), SSN VARCHAR(255))
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 2322 }
What is the day Number and date of all the documents?
SELECT T2.day_Number, T1.Date_Stored FROM All_documents AS T1 JOIN Ref_calendar AS T2 ON T1.Date_Stored = T2.calendar_date
CREATE TABLE All_documents (Date_Stored VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE Ref_calendar (day_Number VARCHAR(255), calendar_date VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_failed", "correction_corrected", "extended", "classified" ]
corrected
1
0
1
{ "split": "train", "row_index": 2423 }
What are the distinct grant amount for the grants where the documents were sent before '1986-08-26 20:49:27' and grant were ended after '1989-03-16 18:27:16'?
SELECT DISTINCT T1.grant_amount FROM Grants AS T1 JOIN Documents AS T2 ON T1.grant_id = T2.grant_id WHERE T2.sent_date < '1986-08-26 20:49:27' AND T1.grant_end_date > '1989-03-16 18:27:16'
CREATE TABLE Grants ( grant_amount VARCHAR(255), grant_id VARCHAR(255), grant_end_date DATETIME ); CREATE TABLE Documents ( grant_id VARCHAR(255), sent_date DATETIME )
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 2493 }
Who is the oldest person?
SELECT name FROM Person WHERE age = (SELECT MAX(age) FROM Person)
CREATE TABLE Person (name VARCHAR(255), age INTEGER)
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
1
1
0
{ "split": "train", "row_index": 2537 }
Who is the oldest person whose job is student?
SELECT name FROM Person WHERE job = 'student' AND age = (SELECT MAX(age) FROM Person WHERE job = 'student')
CREATE TABLE Person ( name VARCHAR(255), job VARCHAR(255), age INTEGER )
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
1
1
1
{ "split": "train", "row_index": 2538 }
Who is the youngest male?
SELECT name FROM Person WHERE gender = 'male' AND age = (SELECT MIN(age) FROM person WHERE gender = 'male')
CREATE TABLE Person (name VARCHAR(255), gender VARCHAR(255), age INTEGER); CREATE TABLE person (name VARCHAR(255), gender VARCHAR(255), age INTEGER)
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
1
1
1
{ "split": "train", "row_index": 2539 }
Find the name of people whose age is greater than any engineer sorted by their age.
SELECT name FROM Person WHERE age > (SELECT MIN(age) FROM Person WHERE job = 'engineer') ORDER BY age
CREATE TABLE Person ( name VARCHAR(255), age INTEGER, job VARCHAR(255) )
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
1
0
{ "split": "train", "row_index": 2548 }
Find the number of people whose age is greater than all engineers.
SELECT COUNT(*) FROM Person WHERE age > (SELECT MAX(age) FROM person WHERE job = 'engineer')
CREATE TABLE Person (age INTEGER, job VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE person (age INTEGER, job VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
1
1
0
{ "split": "train", "row_index": 2549 }
Who has friends that are younger than the average age?
SELECT DISTINCT T2.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T1.age < (SELECT AVG(age) FROM person)
CREATE TABLE person (age INTEGER, id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE Person (name VARCHAR(255), age INTEGER); CREATE TABLE PersonFriend (name VARCHAR(255), friend VARCHAR(255));
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
1
0
{ "split": "train", "row_index": 2565 }
Find Alice's friends of friends.
SELECT DISTINCT T4.name FROM PersonFriend AS T1 JOIN Person AS T2 ON T1.name = T2.name JOIN PersonFriend AS T3 ON T1.friend = T3.name JOIN PersonFriend AS T4 ON T3.friend = T4.name WHERE T2.name = 'Alice' AND T4.name <> 'Alice'
CREATE TABLE PersonFriend (name VARCHAR(255), friend VARCHAR(255)); CREATE TABLE Person (name VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT);
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
4
0
1
{ "split": "train", "row_index": 2574 }
Find names of the document without any images.
SELECT document_name FROM documents EXCEPT SELECT t1.document_name FROM documents AS t1 JOIN document_sections AS t2 ON t1.document_code = t2.document_code JOIN document_sections_images AS t3 ON t2.section_id = t3.section_id
CREATE TABLE document_sections_images (section_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE documents (document_name VARCHAR(255), document_code VARCHAR(255)); CREATE TABLE document_sections (document_code VARCHAR(255), section_id VARCHAR(255))
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 2603 }
Find the name of the department that has no students minored in?
SELECT DName FROM DEPARTMENT WHERE DNO NOT IN (SELECT DNO FROM MINOR_IN)
CREATE TABLE DEPARTMENT (DName VARCHAR(255), DNO VARCHAR(255)); CREATE TABLE MINOR_IN (DNO VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
1
1
0
{ "split": "train", "row_index": 2703 }
Which events have the number of notes between one and three? List the event id and the property id.
SELECT T1.Customer_Event_ID, T1.property_id FROM Customer_Events AS T1 JOIN Customer_Event_Notes AS T2 ON T1.Customer_Event_ID = T2.Customer_Event_ID GROUP BY T1.Customer_Event_ID HAVING COUNT(*) BETWEEN 1 AND 3
CREATE TABLE Customer_Events (Customer_Event_ID VARCHAR(255), property_id VARCHAR(255)); CREATE TABLE Customer_Event_Notes (Customer_Event_ID VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 2795 }
Find the states where have some college students in tryout.
SELECT DISTINCT state FROM college AS T1 JOIN tryout AS T2 ON T1.cName = T2.cName
CREATE TABLE college (cName VARCHAR(255), state VARCHAR(255)); CREATE TABLE tryout (cName VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
mysql
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
1
0
0
{ "split": "train", "row_index": 2870 }
Show the names of all the clients with no booking.
SELECT Customer_Name FROM Clients EXCEPT SELECT T2.Customer_Name FROM Bookings AS T1 JOIN Clients AS T2 ON T1.Customer_ID = T2.Client_ID
CREATE TABLE Clients (Customer_Name VARCHAR(255), Client_ID VARCHAR(255)); CREATE TABLE Bookings (Customer_ID VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT);
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 2954 }
Find all the stage positions of the musicians with first name "Solveig"
SELECT DISTINCT T1.stageposition FROM Performance AS T1 JOIN Band AS T2 ON T1.bandmate = T2.id WHERE T2.Firstname = 'Solveig'
CREATE TABLE Band (id VARCHAR(255), Firstname VARCHAR(255), lastname VARCHAR(255)); CREATE TABLE Performance (stageposition VARCHAR(255), bandmate VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
0
{ "split": "train", "row_index": 2961 }
What is the last name of the musician that has been at the back position the most?
SELECT T2.lastname FROM Performance AS T1 JOIN Band AS T2 ON T1.bandmate = T2.id WHERE T1.stageposition = 'back' GROUP BY T2.lastname ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Band (lastname VARCHAR(255), id VARCHAR(255), firstname VARCHAR(255)); CREATE TABLE Performance (bandmate VARCHAR(255), stageposition VARCHAR(255))
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
5
0
0
{ "split": "train", "row_index": 2971 }
What instrument did the musician with last name "Heilo" use in the song "Le Pop"?
SELECT T4.instrument FROM Performance AS T1 JOIN Band AS T2 ON T1.bandmate = T2.id JOIN Songs AS T3 ON T3.SongId = T1.SongId JOIN Instruments AS T4 ON T4.songid = T3.SongId AND T4.bandmateid = T2.id WHERE T2.lastname = 'Heilo' AND T3.title = 'Le Pop'
CREATE TABLE Songs (SongId VARCHAR(255), title VARCHAR(255)); CREATE TABLE Performance (bandmate VARCHAR(255), SongId VARCHAR(255)); CREATE TABLE Band (id VARCHAR(255), lastname VARCHAR(255), firstname VARCHAR(255)); CREATE TABLE Instruments (instrument VARCHAR(255), songid VARCHAR(255), bandmateid VARCHAR(255))
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
4
0
1
{ "split": "train", "row_index": 2974 }
How many instrument does the musician with last name "Heilo" use?
SELECT COUNT(DISTINCT instrument) FROM instruments AS T1 JOIN Band AS T2 ON T1.bandmateid = T2.id WHERE T2.lastname = 'Heilo'
CREATE TABLE instruments (bandmateid VARCHAR(255), instrument VARCHAR(255)); CREATE TABLE Band (id VARCHAR(255), lastname VARCHAR(255), firstname VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
0
{ "split": "train", "row_index": 2979 }
Find all the instruments ever used by the musician with last name "Heilo"?
SELECT T1.instrument FROM instruments AS T1 JOIN Band AS T2 ON T1.bandmateid = T2.id WHERE T2.lastname = 'Heilo'
CREATE TABLE instruments (bandmateid VARCHAR(255), instrument VARCHAR(255)); CREATE TABLE Band (id VARCHAR(255), lastname VARCHAR(255), firstname VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
0
{ "split": "train", "row_index": 2980 }
Which song has the most vocals?
SELECT T2.title FROM vocals AS T1 JOIN songs AS T2 ON T1.songid = T2.songid GROUP BY T1.songid ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE vocals (songid VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE songs (songid VARCHAR(255), title VARCHAR(255))
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
4
0
0
{ "split": "train", "row_index": 2981 }
Which vocal type has the band mate with last name "Heilo" played the most?
SELECT T1.TYPE FROM vocals AS T1 JOIN band AS T2 ON T1.bandmate = T2.id WHERE T2.lastname = 'Heilo' GROUP BY T1.TYPE ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE vocals (bandmate VARCHAR(255), TYPE VARCHAR(255)); CREATE TABLE band (id VARCHAR(255), lastname VARCHAR(255), firstname VARCHAR(255))
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
5
0
0
{ "split": "train", "row_index": 2983 }
Find the number of vocal types used in song "Demon Kitty Rag"?
SELECT COUNT(*) FROM vocals AS T1 JOIN songs AS T2 ON T1.songid = T2.songid WHERE title = 'Demon Kitty Rag'
CREATE TABLE vocals (songid VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE songs (songid VARCHAR(255), title VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
0
{ "split": "train", "row_index": 2985 }
Which vocal type did the musician with first name "Solveig" played in the song with title "A Bar in Amsterdam"?
SELECT T1.`type` FROM vocals AS T1 JOIN songs AS T2 ON T1.songid = T2.songid JOIN band AS T3 ON T1.bandmate = T3.id WHERE T3.firstname = 'Solveig' AND T2.title = 'A Bar In Amsterdam'
CREATE TABLE vocals (songid VARCHAR(255), bandmate VARCHAR(255), `type` VARCHAR(255)); CREATE TABLE band (id VARCHAR(255), firstname VARCHAR(255), lastname VARCHAR(255)); CREATE TABLE songs (songid VARCHAR(255), title VARCHAR(255))
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_failed", "correction_corrected", "extended", "classified" ]
corrected
3
0
1
{ "split": "train", "row_index": 2987 }
Find the number of vocal types used in song "Le Pop"
SELECT COUNT(*) FROM vocals AS T1 JOIN songs AS T2 ON T1.songid = T2.songid WHERE title = 'Le Pop'
CREATE TABLE vocals (songid VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE songs (songid VARCHAR(255), title VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
0
{ "split": "train", "row_index": 2995 }
Find the first name of the band mate that has performed in most songs.
SELECT t2.firstname FROM Performance AS t1 JOIN Band AS t2 ON t1.bandmate = t2.id JOIN Songs AS t3 ON t3.SongId = t1.SongId GROUP BY t2.firstname ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Performance (bandmate VARCHAR(255), SongId VARCHAR(255)); CREATE TABLE Band (firstname VARCHAR(255), id VARCHAR(255)); CREATE TABLE Songs (SongId VARCHAR(255), title VARCHAR(255))
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_failed", "correction_corrected", "extended", "classified" ]
corrected
5
0
0
{ "split": "train", "row_index": 3000 }
Find the name of companies that do not make DVD drive.
SELECT name FROM manufacturers EXCEPT SELECT T2.name FROM products AS T1 JOIN manufacturers AS T2 ON T1.Manufacturer = T2.code WHERE T1.name = 'DVD drive'
CREATE TABLE manufacturers (name VARCHAR(255), code VARCHAR(255)); CREATE TABLE products (Manufacturer VARCHAR(255), name VARCHAR(255))
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 3029 }
What are the names, details and data types of the characteristics which are never used by any product?
SELECT characteristic_name, other_characteristic_details, characteristic_data_type FROM CHARACTERISTICS EXCEPT SELECT t1.characteristic_name, t1.other_characteristic_details, t1.characteristic_data_type FROM CHARACTERISTICS AS t1 JOIN product_characteristics AS t2 ON t1.characteristic_id = t2.characteristic_id
CREATE TABLE CHARACTERISTICS (characteristic_name VARCHAR(255), other_characteristic_details VARCHAR(255), characteristic_data_type VARCHAR(255), characteristic_id VARCHAR(255)); CREATE TABLE product_characteristics (characteristic_id VARCHAR(255), product_id VARCHAR(255))
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
1
{ "split": "train", "row_index": 3167 }
Find the list of cities that no customer is living in.
SELECT city FROM addresses WHERE NOT city IN (SELECT DISTINCT t3.city FROM customers AS t1 JOIN customer_addresses AS t2 ON t1.customer_id = t2.customer_id JOIN addresses AS t3 ON t2.address_id = t3.address_id)
CREATE TABLE customers (customer_id VARCHAR(255), customer_name VARCHAR(255)); CREATE TABLE addresses (city VARCHAR(255), address_id VARCHAR(255)); CREATE TABLE customer_addresses (customer_id VARCHAR(255), address_id VARCHAR(255));
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
1
1
0
{ "split": "train", "row_index": 3461 }
What is the name of the customer that made the order with the largest quantity?
SELECT t1.customer_name FROM customers AS t1 JOIN customer_orders AS t2 ON t1.customer_id = t2.customer_id JOIN order_items AS t3 ON t2.order_id = t3.order_id WHERE t3.order_quantity = (SELECT MAX(order_quantity) FROM order_items)
CREATE TABLE order_items (order_id VARCHAR(255), order_quantity INTEGER); CREATE TABLE customers (customer_name VARCHAR(255), customer_id VARCHAR(255)); CREATE TABLE customer_orders (customer_id VARCHAR(255), order_id VARCHAR(255))
mysql
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
3
1
0
{ "split": "train", "row_index": 3472 }
Find the names of customers who never ordered product Latte.
SELECT customer_name FROM customers EXCEPT SELECT t1.customer_name FROM customers AS t1 JOIN customer_orders AS t2 ON t1.customer_id = t2.customer_id JOIN order_items AS t3 ON t2.order_id = t3.order_id JOIN products AS t4 ON t3.product_id = t4.product_id WHERE t4.product_details = 'Latte'
CREATE TABLE order_items (order_id VARCHAR(255), product_id VARCHAR(255)); CREATE TABLE customers (customer_name VARCHAR(255), customer_id VARCHAR(255)); CREATE TABLE customer_orders (customer_id VARCHAR(255), order_id VARCHAR(255)); CREATE TABLE products (product_id VARCHAR(255), product_details VARCHAR(255))
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 3487 }
Find the names of customers who never placed an order.
SELECT customer_name FROM customers EXCEPT SELECT t1.customer_name FROM customers AS t1 JOIN customer_orders AS t2 ON t1.customer_id = t2.customer_id
CREATE TABLE customers (customer_name VARCHAR(255), customer_id VARCHAR(255)); CREATE TABLE customer_orders (customer_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
mysql
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
0
1
0
{ "split": "train", "row_index": 3488 }
What are the issue dates of volumes associated with the artist aged 23 or younger?
SELECT Issue_Date FROM artist AS T1 JOIN volume AS T2 ON T1.Artist_ID = T2.Artist_ID WHERE T1.age <= 23
CREATE TABLE volume (Artist_ID VARCHAR(255), Issue_Date DATE); CREATE TABLE artist (Artist_ID VARCHAR(255), age VARCHAR(255))
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
0
{ "split": "train", "row_index": 3517 }
What are the GDP and population of the city that already served as a host more than once?
SELECT t1.gdp, t1.Regional_Population FROM city AS t1 JOIN hosting_city AS t2 ON t1.city_id = t2.host_city GROUP BY t2.host_city HAVING COUNT(*) > 1
CREATE TABLE city (city_id VARCHAR(255), gdp INTEGER, Regional_Population INTEGER); CREATE TABLE hosting_city (host_city VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
mysql
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validation_failed", "correction_failed", "correction_corrected", "extended", "classified" ]
corrected
2
0
1
{ "split": "train", "row_index": 3577 }