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
List Aerosmith's albums.
SELECT T1.title FROM albums AS T1 JOIN artists AS T2 ON T1.artist_id = T2.id WHERE T2.name = 'Aerosmith'
CREATE TABLE artists (id VARCHAR(255), name VARCHAR(255)); CREATE TABLE albums (title VARCHAR(255), artist_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 316 }
How many albums does Billy Cobham has?
SELECT COUNT(*) FROM albums AS T1 JOIN artists AS T2 ON T1.artist_id = T2.id WHERE T2.name = 'Billy Cobham'
CREATE TABLE artists (id VARCHAR(255), name VARCHAR(255)); CREATE TABLE albums (artist_id VARCHAR(255), title VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 317 }
Eduardo Martins is a customer at which company?
SELECT company FROM customers WHERE first_name = 'Eduardo' AND last_name = 'Martins'
CREATE TABLE customers (company VARCHAR(255), first_name VARCHAR(255), last_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 318 }
What is Astrid Gruber's email and phone number?
SELECT email, phone FROM customers WHERE first_name = 'Astrid' AND last_name = 'Gruber'
CREATE TABLE customers (email VARCHAR(255), phone VARCHAR(255), first_name VARCHAR(255), last_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
2
{ "split": "train", "row_index": 319 }
How many customers live in Prague city?
SELECT COUNT(*) FROM customers WHERE city = 'Prague'
CREATE TABLE customers (city VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 320 }
How many customers in state of CA?
SELECT COUNT(*) FROM customers WHERE state = 'CA'
CREATE TABLE customers (state VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 321 }
What country does Roberto Almeida live?
SELECT country FROM customers WHERE first_name = 'Roberto' AND last_name = 'Almeida'
CREATE TABLE customers (country VARCHAR(255), first_name VARCHAR(255), last_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 322 }
List the name of albums that are released by aritist whose name has 'Led'
SELECT T2.title FROM artists AS T1 JOIN albums AS T2 ON T1.id = T2.artist_id WHERE T1.name LIKE '%Led%'
CREATE TABLE artists (id VARCHAR(255), name VARCHAR(255)); CREATE TABLE albums (title VARCHAR(255), artist_id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 323 }
How many customers does Steve Johnson support?
SELECT COUNT(*) FROM employees AS T1 JOIN customers AS T2 ON T2.support_rep_id = T1.id WHERE T1.first_name = 'Steve' AND T1.last_name = 'Johnson'
CREATE TABLE employees (id VARCHAR(255), first_name VARCHAR(255), last_name VARCHAR(255)); CREATE TABLE customers (support_rep_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 324 }
What is the title, phone and hire date of Nancy Edwards?
SELECT title, phone, hire_date FROM employees WHERE first_name = 'Nancy' AND last_name = 'Edwards'
CREATE TABLE employees (title VARCHAR(255), phone VARCHAR(255), hire_date VARCHAR(255), first_name VARCHAR(255), last_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
2
{ "split": "train", "row_index": 325 }
What is the address of employee Nancy Edwards?
SELECT address FROM employees WHERE first_name = 'Nancy' AND last_name = 'Edwards'
CREATE TABLE employees (address VARCHAR(255), first_name VARCHAR(255), last_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 327 }
Find the full name of employee who supported the most number of customers.
SELECT T1.first_name, T1.last_name FROM employees AS T1 JOIN customers AS T2 ON T1.id = T2.support_rep_id GROUP BY T1.id ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE employees (first_name VARCHAR(255), last_name VARCHAR(255), id VARCHAR(255)); CREATE TABLE customers (support_rep_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 328 }
How many employees are living in Canada?
SELECT COUNT(*) FROM employees WHERE country = 'Canada'
CREATE TABLE employees (country VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 329 }
What is employee Nancy Edwards's phone number?
SELECT phone FROM employees WHERE first_name = 'Nancy' AND last_name = 'Edwards'
CREATE TABLE employees (phone VARCHAR(255), first_name VARCHAR(255), last_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 330 }
Who is the youngest employee in the company? List employee's first and last name.
SELECT first_name, last_name FROM employees ORDER BY birth_date DESC LIMIT 1
CREATE TABLE employees (first_name VARCHAR(255), last_name VARCHAR(255), birth_date VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 331 }
List top 10 employee work longest in the company. List employee's first and last name.
SELECT first_name, last_name FROM employees ORDER BY hire_date LIMIT 10
CREATE TABLE employees (first_name VARCHAR(255), last_name VARCHAR(255), hire_date VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 332 }
Find the number of employees whose title is IT Staff from each city?
SELECT COUNT(*), city FROM employees WHERE title = 'IT Staff' GROUP BY city
CREATE TABLE employees (city VARCHAR(255), title VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 333 }
How many orders does Lucas Mancini has?
SELECT COUNT(*) FROM customers AS T1 JOIN invoices AS T2 ON T1.id = T2.customer_id WHERE T1.first_name = 'Lucas' AND T1.last_name = 'Mancini'
CREATE TABLE invoices (customer_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE customers (id VARCHAR(255), first_name VARCHAR(255), last_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 335 }
What is the total amount of money spent by Lucas Mancini?
SELECT SUM(T2.total) FROM customers AS T1 JOIN invoices AS T2 ON T1.id = T2.customer_id WHERE T1.first_name = 'Lucas' AND T1.last_name = 'Mancini'
CREATE TABLE invoices (total INTEGER, customer_id VARCHAR(255)); CREATE TABLE customers (id VARCHAR(255), first_name VARCHAR(255), last_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 336 }
List all media types.
SELECT name FROM media_types
CREATE TABLE media_types (name VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 337 }
List all different genre types.
SELECT DISTINCT name FROM genres
CREATE TABLE genres (name VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 338 }
List the name of all playlist.
SELECT name FROM playlists
CREATE TABLE playlists (name VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 339 }
Who is the composer of track Fast As a Shark?
SELECT composer FROM tracks WHERE name = 'Fast As a Shark'
CREATE TABLE tracks (composer VARCHAR(255), name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 340 }
How long does track Fast As a Shark has?
SELECT milliseconds FROM tracks WHERE name = 'Fast As a Shark'
CREATE TABLE tracks (milliseconds VARCHAR(255), name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 341 }
What is the name of tracks whose genre is Rock?
SELECT T2.name FROM genres AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id WHERE T1.name = 'Rock'
CREATE TABLE genres (id VARCHAR(255), name VARCHAR(255)); CREATE TABLE tracks (name VARCHAR(255), genre_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 342 }
What is title of album which track Balls to the Wall belongs to?
SELECT T1.title FROM albums AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id WHERE T2.name = 'Balls to the Wall'
CREATE TABLE tracks (genre_id VARCHAR(255), name VARCHAR(255)); CREATE TABLE albums (title VARCHAR(255), id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 343 }
List name of all tracks in Balls to the Wall.
SELECT T2.name FROM albums AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id WHERE T1.title = 'Balls to the Wall'
CREATE TABLE tracks (name VARCHAR(255), genre_id VARCHAR(255)); CREATE TABLE albums (id VARCHAR(255), title VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 344 }
List title of albums have the number of tracks greater than 10.
SELECT T1.title FROM albums AS T1 JOIN tracks AS T2 ON T1.id = T2.album_id GROUP BY T1.id HAVING COUNT(T1.id) > 10
CREATE TABLE tracks (album_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE albums (title VARCHAR(255), id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 345 }
List the name of tracks belongs to genre Rock and whose media type is MPEG audio file.
SELECT T2.name FROM genres AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id JOIN media_types AS T3 ON T3.id = T2.media_type_id WHERE T1.name = 'Rock' AND T3.name = 'MPEG audio file'
CREATE TABLE genres (id VARCHAR(255), name VARCHAR(255)); CREATE TABLE tracks (name VARCHAR(255), genre_id VARCHAR(255), media_type_id VARCHAR(255)); CREATE TABLE media_types (id VARCHAR(255), name VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 346 }
List the name of tracks belongs to genre Rock or media type is MPEG audio file.
SELECT T2.name FROM genres AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id JOIN media_types AS T3 ON T3.id = T2.media_type_id WHERE T1.name = 'Rock' OR T3.name = 'MPEG audio file'
CREATE TABLE genres (id VARCHAR(255), name VARCHAR(255)); CREATE TABLE tracks (name VARCHAR(255), genre_id VARCHAR(255), media_type_id VARCHAR(255)); CREATE TABLE media_types (id VARCHAR(255), name VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 347 }
List the name of tracks belongs to genre Rock or genre Jazz.
SELECT T2.name FROM genres AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id WHERE T1.name = 'Rock' OR T1.name = 'Jazz'
CREATE TABLE genres (id VARCHAR(255), name VARCHAR(255)); CREATE TABLE tracks (name VARCHAR(255), genre_id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 348 }
List the name of all tracks in the playlists of Movies.
SELECT T1.name FROM tracks AS T1 JOIN playlist_tracks AS T2 ON T1.id = T2.track_id JOIN playlists AS T3 ON T3.id = T2.playlist_id WHERE T3.name = 'Movies'
CREATE TABLE playlists (id VARCHAR(255), name VARCHAR(255)); CREATE TABLE playlist_tracks (track_id VARCHAR(255), playlist_id VARCHAR(255)); CREATE TABLE tracks (name VARCHAR(255), id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 349 }
List the name of playlist which has number of tracks greater than 100.
SELECT T2.name FROM playlist_tracks AS T1 JOIN playlists AS T2 ON T2.id = T1.playlist_id GROUP BY T1.playlist_id HAVING COUNT(T1.track_id) > 100
CREATE TABLE playlist_tracks (playlist_id VARCHAR(255), track_id VARCHAR(255)); CREATE TABLE playlists (name VARCHAR(255), id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 350 }
List all tracks bought by customer Daan Peeters.
SELECT T1.name FROM tracks AS T1 JOIN invoice_lines AS T2 ON T1.id = T2.track_id JOIN invoices AS T3 ON T3.id = T2.invoice_id JOIN customers AS T4 ON T4.id = T3.customer_id WHERE T4.first_name = 'Daan' AND T4.last_name = 'Peeters'
CREATE TABLE invoices (id VARCHAR(255), customer_id VARCHAR(255)); CREATE TABLE invoice_lines (track_id VARCHAR(255), invoice_id VARCHAR(255)); CREATE TABLE tracks (name VARCHAR(255), id VARCHAR(255)); CREATE TABLE customers (id VARCHAR(255), first_name VARCHAR(255), last_name VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 351 }
How much is the track Fast As a Shark?
SELECT unit_price FROM tracks WHERE name = 'Fast As a Shark'
CREATE TABLE tracks (unit_price VARCHAR(255), name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 352 }
Find the name of tracks which are in Movies playlist but not in music playlist.
SELECT T1.name FROM tracks AS T1 JOIN playlist_tracks AS T2 ON T1.id = T2.track_id JOIN playlists AS T3 ON T2.playlist_id = T3.id WHERE T3.name = 'Movies' EXCEPT SELECT T1.name FROM tracks AS T1 JOIN playlist_tracks AS T2 ON T1.id = T2.track_id JOIN playlists AS T3 ON T2.playlist_id = T3.id WHERE T3.name = 'Music'
CREATE TABLE playlists (id VARCHAR(255), name VARCHAR(255)); CREATE TABLE playlist_tracks (track_id VARCHAR(255), playlist_id VARCHAR(255)); CREATE TABLE tracks (name VARCHAR(255), id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
1
0
{ "split": "train", "row_index": 353 }
Find the name of tracks which are in both Movies and music playlists.
SELECT T1.name FROM tracks AS T1 JOIN playlist_tracks AS T2 ON T1.id = T2.track_id JOIN playlists AS T3 ON T2.playlist_id = T3.id WHERE T3.name = 'Movies' INTERSECT SELECT T1.name FROM tracks AS T1 JOIN playlist_tracks AS T2 ON T1.id = T2.track_id JOIN playlists AS T3 ON T2.playlist_id = T3.id WHERE T3.name = 'Music'
CREATE TABLE playlists (id VARCHAR(255), name VARCHAR(255)); CREATE TABLE playlist_tracks (track_id VARCHAR(255), playlist_id VARCHAR(255)); CREATE TABLE tracks (name VARCHAR(255), id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
1
0
{ "split": "train", "row_index": 354 }
Find number of tracks in each genre?
SELECT COUNT(*), T1.name FROM genres AS T1 JOIN tracks AS T2 ON T1.id = T2.genre_id GROUP BY T1.name
CREATE TABLE tracks (genre_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE genres (name VARCHAR(255), id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 355 }
How many editors are there?
SELECT COUNT(*) FROM editor
CREATE TABLE editor (Id VARCHAR(255), Name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 356 }
List the names of editors in ascending order of age.
SELECT Name FROM editor ORDER BY Age
CREATE TABLE editor (Name VARCHAR(255), Age VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 357 }
What are the names and ages of editors?
SELECT Name, Age FROM editor
CREATE TABLE editor (Name VARCHAR(255), Age VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 358 }
List the names of editors who are older than 25.
SELECT Name FROM editor WHERE Age > 25
CREATE TABLE editor (Name VARCHAR(255), Age INTEGER)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 359 }
Show the names of editors of age either 24 or 25.
SELECT Name FROM editor WHERE Age = 24 OR Age = 25
CREATE TABLE editor (Name VARCHAR(255), Age VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 360 }
What is the name of the youngest editor?
SELECT Name FROM editor ORDER BY Age LIMIT 1
CREATE TABLE editor (Name VARCHAR(255), Age VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 361 }
What are the different ages of editors? Show each age along with the number of editors of that age.
SELECT Age, COUNT(*) FROM editor GROUP BY Age
CREATE TABLE editor (Age VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 362 }
Please show the most common age of editors.
SELECT Age FROM editor GROUP BY Age ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE editor (Age VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 363 }
Show the distinct themes of journals.
SELECT DISTINCT Theme FROM journal
CREATE TABLE journal (Theme VARCHAR(255), Journal_ID VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 364 }
Show the names of editors and the theme of journals for which they serve on committees.
SELECT T2.Name, T3.Theme FROM journal_committee AS T1 JOIN editor AS T2 ON T1.Editor_ID = T2.Editor_ID JOIN journal AS T3 ON T1.Journal_ID = T3.Journal_ID
CREATE TABLE journal_committee (Editor_ID VARCHAR(255), Journal_ID VARCHAR(255)); CREATE TABLE editor (Name VARCHAR(255), Editor_ID VARCHAR(255)); CREATE TABLE journal (Theme VARCHAR(255), Journal_ID VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 365 }
Show the names and ages of editors and the theme of journals for which they serve on committees, in ascending alphabetical order of theme.
SELECT T2.Name, T2.age, T3.Theme FROM journal_committee AS T1 JOIN editor AS T2 ON T1.Editor_ID = T2.Editor_ID JOIN journal AS T3 ON T1.Journal_ID = T3.Journal_ID ORDER BY T3.Theme
CREATE TABLE journal_committee (Editor_ID VARCHAR(255), Journal_ID VARCHAR(255)); CREATE TABLE editor (Name VARCHAR(255), age VARCHAR(255), Editor_ID VARCHAR(255)); CREATE TABLE journal (Theme VARCHAR(255), Journal_ID VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 366 }
Show the names of editors that are on the committee of journals with sales bigger than 3000.
SELECT T2.Name FROM journal_committee AS T1 JOIN editor AS T2 ON T1.Editor_ID = T2.Editor_ID JOIN journal AS T3 ON T1.Journal_ID = T3.Journal_ID WHERE T3.Sales > 3000
CREATE TABLE journal_committee (Editor_ID VARCHAR(255), Journal_ID VARCHAR(255)); CREATE TABLE editor (Name VARCHAR(255), Editor_ID VARCHAR(255)); CREATE TABLE journal (Journal_ID VARCHAR(255), Sales INTEGER)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 367 }
Show the names of editors that are on at least two journal committees.
SELECT T1.Name FROM editor AS T1 JOIN journal_committee AS T2 ON T1.Editor_ID = T2.Editor_ID GROUP BY T1.Name HAVING COUNT(*) >= 2
CREATE TABLE editor (Name VARCHAR(255), Editor_ID VARCHAR(255)); CREATE TABLE journal_committee (Editor_ID VARCHAR(255), Journal_ID VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 369 }
List the names of editors that are not on any journal committee.
SELECT Name FROM editor WHERE NOT editor_id IN (SELECT editor_id FROM journal_committee)
CREATE TABLE editor (Name VARCHAR(255), editor_id VARCHAR(255)); CREATE TABLE journal_committee (Name VARCHAR(255), editor_id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 370 }
What is the average sales of the journals that have an editor whose work type is 'Photo'?
SELECT AVG(T1.sales) FROM journal AS T1 JOIN journal_committee AS T2 ON T1.journal_ID = T2.journal_ID WHERE T2.work_type = 'Photo'
CREATE TABLE journal_committee (journal_ID VARCHAR(255), work_type VARCHAR(255)); CREATE TABLE journal (sales INTEGER, journal_ID VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 372 }
How many accounts do we have?
SELECT COUNT(*) FROM Accounts
CREATE TABLE Accounts (Id VARCHAR(255), account_name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 373 }
Show ids, customer ids, names for all accounts.
SELECT account_id, customer_id, account_name FROM Accounts
CREATE TABLE Accounts (account_id VARCHAR(255), customer_id VARCHAR(255), account_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 374 }
Show other account details for account with name 338.
SELECT other_account_details FROM Accounts WHERE account_name = '338'
CREATE TABLE Accounts (other_account_details VARCHAR(255), account_name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 375 }
What is the first name, last name, and phone of the customer with account name 162?
SELECT T2.customer_first_name, T2.customer_last_name, T2.customer_phone FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T1.account_name = '162'
CREATE TABLE Customers (customer_first_name VARCHAR(255), customer_last_name VARCHAR(255), customer_phone VARCHAR(255), customer_id VARCHAR(255)); CREATE TABLE Accounts (customer_id VARCHAR(255), account_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 376 }
How many accounts does the customer with first name Art and last name Turcotte have?
SELECT COUNT(*) FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.customer_first_name = 'Art' AND T2.customer_last_name = 'Turcotte'
CREATE TABLE Accounts (customer_id VARCHAR(255), account_id VARCHAR(255)); CREATE TABLE Customers (customer_id VARCHAR(255), customer_first_name VARCHAR(255), customer_last_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 377 }
Show all customer ids and the number of accounts for each customer.
SELECT customer_id, COUNT(*) FROM Accounts GROUP BY customer_id
CREATE TABLE Accounts (customer_id VARCHAR(255), account_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 378 }
Show the customer id and number of accounts with most accounts.
SELECT customer_id, COUNT(*) FROM Accounts GROUP BY customer_id ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Accounts (customer_id VARCHAR(255), account_number VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
2
{ "split": "train", "row_index": 379 }
What is the customer first, last name and id with least number of accounts.
SELECT T2.customer_first_name, T2.customer_last_name, T1.customer_id FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY COUNT(*) LIMIT 1
CREATE TABLE Accounts (customer_id VARCHAR(255), account_type VARCHAR(255)); CREATE TABLE Customers (customer_first_name VARCHAR(255), customer_last_name VARCHAR(255), customer_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 380 }
Show the number of all customers without an account.
SELECT COUNT(*) FROM Customers WHERE NOT customer_id IN (SELECT customer_id FROM Accounts)
CREATE TABLE Customers (customer_id VARCHAR(255), customer_name VARCHAR(255)); CREATE TABLE Accounts (customer_id VARCHAR(255), account_number VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
1
{ "split": "train", "row_index": 381 }
Show distinct first and last names for all customers with an account.
SELECT DISTINCT T1.customer_first_name, T1.customer_last_name FROM Customers AS T1 JOIN Accounts AS T2 ON T1.customer_id = T2.customer_id
CREATE TABLE Accounts (customer_id VARCHAR(255), account_type VARCHAR(255)); CREATE TABLE Customers (customer_first_name VARCHAR(255), customer_last_name VARCHAR(255), customer_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 383 }
How many customers have an account?
SELECT COUNT(DISTINCT customer_id) FROM Accounts
CREATE TABLE Accounts (customer_id VARCHAR(255), account_status VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 384 }
How many customers do we have?
SELECT COUNT(*) FROM Customers
CREATE TABLE Customers (Id VARCHAR(255), name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 385 }
Show ids, first names, last names, and phones for all customers.
SELECT customer_id, customer_first_name, customer_last_name, customer_phone FROM Customers
CREATE TABLE Customers (customer_id VARCHAR(255), customer_first_name VARCHAR(255), customer_last_name VARCHAR(255), customer_phone VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 386 }
What is the phone and email for customer with first name Aniyah and last name Feest?
SELECT customer_phone, customer_email FROM Customers WHERE customer_first_name = 'Aniyah' AND customer_last_name = 'Feest'
CREATE TABLE Customers (customer_phone VARCHAR(255), customer_email VARCHAR(255), customer_first_name VARCHAR(255), customer_last_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
2
{ "split": "train", "row_index": 387 }
Show the number of customer cards.
SELECT COUNT(*) FROM Customers_cards
CREATE TABLE Customers_cards (Id VARCHAR(255), card_number VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 388 }
Show ids, customer ids, card type codes, card numbers for all cards.
SELECT card_id, customer_id, card_type_code, card_number FROM Customers_cards
CREATE TABLE Customers_cards (card_id VARCHAR(255), customer_id VARCHAR(255), card_type_code VARCHAR(255), card_number VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 389 }
Show the date valid from and the date valid to for the card with card number '4560596484842'.
SELECT date_valid_from, date_valid_to FROM Customers_cards WHERE card_number = '4560596484842'
CREATE TABLE Customers_cards (date_valid_from VARCHAR(255), date_valid_to VARCHAR(255), card_number VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 390 }
What is the first name, last name, and phone of the customer with card 4560596484842.
SELECT T2.customer_first_name, T2.customer_last_name, T2.customer_phone FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T1.card_number = '4560596484842'
CREATE TABLE Customers (customer_first_name VARCHAR(255), customer_last_name VARCHAR(255), customer_phone VARCHAR(255), customer_id VARCHAR(255)); CREATE TABLE Customers_cards (customer_id VARCHAR(255), card_number VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 391 }
How many cards does customer Art Turcotte have?
SELECT COUNT(*) FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.customer_first_name = 'Art' AND T2.customer_last_name = 'Turcotte'
CREATE TABLE Customers_cards (customer_id VARCHAR(255), card_number VARCHAR(255)); CREATE TABLE Customers (customer_id VARCHAR(255), customer_first_name VARCHAR(255), customer_last_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 392 }
How many debit cards do we have?
SELECT COUNT(*) FROM Customers_cards WHERE card_type_code = 'Debit'
CREATE TABLE Customers_cards (card_type_code VARCHAR(255), card_number VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 393 }
How many credit cards does customer Blanche Huels have?
SELECT COUNT(*) FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.customer_first_name = 'Blanche' AND T2.customer_last_name = 'Huels' AND T1.card_type_code = 'Credit'
CREATE TABLE Customers_cards (customer_id VARCHAR(255), card_type_code VARCHAR(255)); CREATE TABLE Customers (customer_id VARCHAR(255), customer_first_name VARCHAR(255), customer_last_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 394 }
Show all customer ids and the number of cards owned by each customer.
SELECT customer_id, COUNT(*) FROM Customers_cards GROUP BY customer_id
CREATE TABLE Customers_cards (customer_id VARCHAR(255), card_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 395 }
What is the customer id with most number of cards, and how many does he have?
SELECT customer_id, COUNT(*) FROM Customers_cards GROUP BY customer_id ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Customers_cards (customer_id VARCHAR(255), card_id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
2
{ "split": "train", "row_index": 396 }
Show id, first and last names for all customers with at least two cards.
SELECT T1.customer_id, T2.customer_first_name, T2.customer_last_name FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id HAVING COUNT(*) >= 2
CREATE TABLE Customers_cards (customer_id VARCHAR(255), card_id VARCHAR(255)); CREATE TABLE Customers (customer_first_name VARCHAR(255), customer_last_name VARCHAR(255), customer_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 397 }
What is the customer id, first and last name with least number of accounts.
SELECT T1.customer_id, T2.customer_first_name, T2.customer_last_name FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY COUNT(*) LIMIT 1
CREATE TABLE Customers_cards (customer_id VARCHAR(255), card_id VARCHAR(255)); CREATE TABLE Customers (customer_first_name VARCHAR(255), customer_last_name VARCHAR(255), customer_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 398 }
Show all card type codes and the number of cards in each type.
SELECT card_type_code, COUNT(*) FROM Customers_cards GROUP BY card_type_code
CREATE TABLE Customers_cards (card_type_code VARCHAR(255), card_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 399 }
What is the card type code with most number of cards?
SELECT card_type_code FROM Customers_cards GROUP BY card_type_code ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Customers_cards (card_type_code VARCHAR(255), card_id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 400 }
Show card type codes with at least 5 cards.
SELECT card_type_code FROM Customers_cards GROUP BY card_type_code HAVING COUNT(*) >= 5
CREATE TABLE Customers_cards (card_type_code VARCHAR(255), card_id VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 401 }
Show all card type codes and the number of customers holding cards in each type.
SELECT card_type_code, COUNT(DISTINCT customer_id) FROM Customers_cards GROUP BY card_type_code
CREATE TABLE Customers_cards (card_type_code VARCHAR(255), customer_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 402 }
Show all card type codes.
SELECT DISTINCT card_type_code FROM Customers_Cards
CREATE TABLE Customers_Cards (card_type_code VARCHAR(255), card_id VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 404 }
Show the number of card types.
SELECT COUNT(DISTINCT card_type_code) FROM Customers_Cards
CREATE TABLE Customers_Cards (card_type_code VARCHAR(255), card_id VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 405 }
Show all transaction types.
SELECT DISTINCT transaction_type FROM Financial_Transactions
CREATE TABLE Financial_Transactions (transaction_type VARCHAR(255), transaction_amount INTEGER)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 406 }
Show the number of transaction types.
SELECT COUNT(DISTINCT transaction_type) FROM Financial_Transactions
CREATE TABLE Financial_Transactions (transaction_type VARCHAR(255), transaction_amount INTEGER)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 407 }
What is the average and total transaction amount?
SELECT AVG(transaction_amount), SUM(transaction_amount) FROM Financial_transactions
CREATE TABLE Financial_transactions (transaction_amount INTEGER, transaction_type VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
2
{ "split": "train", "row_index": 408 }
Show the card type codes and the number of transactions.
SELECT T2.card_type_code, COUNT(*) FROM Financial_transactions AS T1 JOIN Customers_cards AS T2 ON T1.card_id = T2.card_id GROUP BY T2.card_type_code
CREATE TABLE Financial_transactions (card_id VARCHAR(255), transaction_amount INTEGER); CREATE TABLE Customers_cards (card_type_code VARCHAR(255), card_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 409 }
Show the transaction type and the number of transactions.
SELECT transaction_type, COUNT(*) FROM Financial_transactions GROUP BY transaction_type
CREATE TABLE Financial_transactions (transaction_type VARCHAR(255), transaction_amount INTEGER)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 410 }
What is the transaction type that has processed the greatest total amount in transactions?
SELECT transaction_type FROM Financial_transactions GROUP BY transaction_type ORDER BY SUM(transaction_amount) DESC LIMIT 1
CREATE TABLE Financial_transactions (transaction_type VARCHAR(255), transaction_amount INTEGER)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 411 }
Show the account id and the number of transactions for each account
SELECT account_id, COUNT(*) FROM Financial_transactions GROUP BY account_id
CREATE TABLE Financial_transactions (account_id VARCHAR(255), transaction_amount INTEGER)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 412 }
How many tracks do we have?
SELECT COUNT(*) FROM track
CREATE TABLE track (Id VARCHAR(255), name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 413 }
Show the name and location for all tracks.
SELECT name, LOCATION FROM track
CREATE TABLE track (name VARCHAR(255), LOCATION VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 414 }
Show names and seatings, ordered by seating for all tracks opened after 2000.
SELECT name, seating FROM track WHERE year_opened > 2000 ORDER BY seating
CREATE TABLE track (name VARCHAR(255), seating VARCHAR(255), year_opened INTEGER)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 415 }
What is the name, location and seating for the most recently opened track?
SELECT name, LOCATION, seating FROM track ORDER BY year_opened DESC LIMIT 1
CREATE TABLE track (name VARCHAR(255), LOCATION VARCHAR(255), seating VARCHAR(255), year_opened VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 416 }
What is the minimum, maximum, and average seating for all tracks.
SELECT MIN(seating), MAX(seating), AVG(seating) FROM track
CREATE TABLE track (seating INTEGER, id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
2
{ "split": "train", "row_index": 417 }
Show the name, location, open year for all tracks with a seating higher than the average.
SELECT name, LOCATION, year_opened FROM track WHERE seating > (SELECT AVG(seating) FROM track)
CREATE TABLE track (name VARCHAR(255), LOCATION VARCHAR(255), year_opened VARCHAR(255), seating INTEGER)
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
1
{ "split": "train", "row_index": 418 }
What are distinct locations where tracks are located?
SELECT DISTINCT LOCATION FROM track
CREATE TABLE track (LOCATION VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 419 }
How many races are there?
SELECT COUNT(*) FROM race
CREATE TABLE race (Id VARCHAR(255), name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 420 }
What are the distinct classes that races can have?
SELECT DISTINCT CLASS FROM race
CREATE TABLE race (CLASS VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 421 }