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
What is the description of the claim status "Open"?
SELECT claim_status_description FROM claims_processing_stages WHERE claim_status_name = 'Open'
CREATE TABLE claims_processing_stages (claim_status_description VARCHAR(255), claim_status_name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 883 }
How many distinct claim outcome codes are there?
SELECT COUNT(DISTINCT claim_outcome_code) FROM claims_processing
CREATE TABLE claims_processing (claim_outcome_code VARCHAR(255), id INTEGER PRIMARY KEY AUTOINCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 884 }
Show the id, the date of account opened, the account name, and other account detail for all accounts.
SELECT account_id, date_account_opened, account_name, other_account_details FROM Accounts
CREATE TABLE Accounts (account_id VARCHAR(255), date_account_opened VARCHAR(255), account_name VARCHAR(255), other_account_details VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 886 }
Show the id, the account name, and other account details for all accounts by the customer with first name 'Meaghan'.
SELECT T1.account_id, T1.date_account_opened, T1.account_name, T1.other_account_details FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.customer_first_name = 'Meaghan'
CREATE TABLE Accounts (account_id VARCHAR(255), date_account_opened VARCHAR(255), account_name VARCHAR(255), other_account_details VARCHAR(255), customer_id VARCHAR(255)); CREATE TABLE Customers (customer_id VARCHAR(255), customer_first_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 887 }
Show the account name and other account detail for all accounts by the customer with first name Meaghan and last name Keeling.
SELECT T1.account_name, T1.other_account_details FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.customer_first_name = 'Meaghan' AND T2.customer_last_name = 'Keeling'
CREATE TABLE Customers (customer_id VARCHAR(255), customer_first_name VARCHAR(255), customer_last_name VARCHAR(255)); CREATE TABLE Accounts (account_name VARCHAR(255), other_account_details VARCHAR(255), customer_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
2
{ "split": "train", "row_index": 888 }
Show the first name and last name for the customer with account name 900.
SELECT T2.customer_first_name, T2.customer_last_name FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T1.account_name = '900'
CREATE TABLE Accounts (customer_id VARCHAR(255), account_name 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": 889 }
Show the unique first names, last names, and phone numbers for all customers with any account.
SELECT DISTINCT T1.customer_first_name, T1.customer_last_name, T1.phone_number FROM Customers AS T1 JOIN Accounts AS T2 ON T1.customer_id = T2.customer_id
CREATE TABLE Accounts (customer_id VARCHAR(255), id INTEGER PRIMARY KEY AUTOINCREMENT); CREATE TABLE Customers (customer_first_name VARCHAR(255), customer_last_name VARCHAR(255), phone_number VARCHAR(255), customer_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 890 }
Show customer ids who don't have an account.
SELECT customer_id FROM Customers EXCEPT SELECT customer_id FROM Accounts
CREATE TABLE Customers (customer_id VARCHAR(255), name VARCHAR(255)); CREATE TABLE Accounts (customer_id VARCHAR(255), account_type VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
1
0
{ "split": "train", "row_index": 891 }
How many accounts does each customer have? List the number and customer id.
SELECT COUNT(*), customer_id FROM Accounts GROUP BY customer_id
CREATE TABLE Accounts (customer_id VARCHAR(255), account_type VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 892 }
What is the customer id, first and last name with most number of accounts.
SELECT T1.customer_id, T2.customer_first_name, T2.customer_last_name FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Accounts (customer_id VARCHAR(255), account_status 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": 893 }
Show id, first name and last name for all customers and the number of accounts.
SELECT T1.customer_id, T2.customer_first_name, T2.customer_last_name, COUNT(*) FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id
CREATE TABLE Accounts (customer_id VARCHAR(255), account_status 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": 894 }
Show first name and id for all customers with at least 2 accounts.
SELECT T2.customer_first_name, T1.customer_id FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id HAVING COUNT(*) >= 2
CREATE TABLE Customers (customer_first_name VARCHAR(255), customer_id VARCHAR(255), customer_last_name VARCHAR(255)); CREATE TABLE Accounts (customer_id VARCHAR(255), account_type VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 895 }
Show the number of customers for each gender.
SELECT gender, COUNT(*) FROM Customers GROUP BY gender
CREATE TABLE Customers (gender VARCHAR(255), customer_id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 896 }
How many transactions do we have?
SELECT COUNT(*) FROM Financial_transactions
CREATE TABLE Financial_transactions (Id VARCHAR(255), transaction_amount INTEGER)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 897 }
How many transaction does each account have? Show the number and account id.
SELECT COUNT(*), account_id FROM Financial_transactions
CREATE TABLE Financial_transactions (account_id VARCHAR(255), transaction_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 898 }
How many transaction does account with name 337 have?
SELECT COUNT(*) FROM Financial_transactions AS T1 JOIN Accounts AS T2 ON T1.account_id = T2.account_id WHERE T2.account_name = '337'
CREATE TABLE Accounts (account_id VARCHAR(255), account_name VARCHAR(255)); CREATE TABLE Financial_transactions (account_id VARCHAR(255), transaction_amount INTEGER)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 899 }
What is the average, minimum, maximum, and total transaction amount?
SELECT AVG(transaction_amount), MIN(transaction_amount), MAX(transaction_amount), SUM(transaction_amount) FROM Financial_transactions
CREATE TABLE Financial_transactions (transaction_amount INTEGER, transaction_id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
2
{ "split": "train", "row_index": 900 }
Show ids for all transactions whose amounts are greater than the average.
SELECT transaction_id FROM Financial_transactions WHERE transaction_amount > (SELECT AVG(transaction_amount) FROM Financial_transactions)
CREATE TABLE Financial_transactions (transaction_id VARCHAR(255), transaction_amount INTEGER)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 901 }
Show the transaction types and the total amount of transactions.
SELECT transaction_type, SUM(transaction_amount) 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": 902 }
Show the account name, id and the number of transactions for each account.
SELECT T2.account_name, T1.account_id, COUNT(*) FROM Financial_transactions AS T1 JOIN Accounts AS T2 ON T1.account_id = T2.account_id GROUP BY T1.account_id
CREATE TABLE Financial_transactions (account_id VARCHAR(255), transaction_id VARCHAR(255)); CREATE TABLE Accounts (account_name VARCHAR(255), account_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 903 }
Show the account id with most number of transactions.
SELECT account_id FROM Financial_transactions GROUP BY account_id ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Financial_transactions (account_id VARCHAR(255), transaction_id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 904 }
Show the account id and name with at least 4 transactions.
SELECT T1.account_id, T2.account_name FROM Financial_transactions AS T1 JOIN Accounts AS T2 ON T1.account_id = T2.account_id GROUP BY T1.account_id HAVING COUNT(*) >= 4
CREATE TABLE Financial_transactions (account_id VARCHAR(255), transaction_amount INTEGER); CREATE TABLE Accounts (account_name VARCHAR(255), account_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 905 }
Show all product sizes.
SELECT DISTINCT product_size FROM Products
CREATE TABLE Products (product_size VARCHAR(255), product_id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 906 }
Show all product colors.
SELECT DISTINCT product_color FROM Products
CREATE TABLE Products (product_color 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": 907 }
Show the invoice number and the number of transactions for each invoice.
SELECT invoice_number, COUNT(*) FROM Financial_transactions GROUP BY invoice_number
CREATE TABLE Financial_transactions (invoice_number VARCHAR(255), transaction_id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 908 }
What is the invoice number and invoice date for the invoice with most number of transactions?
SELECT T2.invoice_number, T2.invoice_date FROM Financial_transactions AS T1 JOIN Invoices AS T2 ON T1.invoice_number = T2.invoice_number GROUP BY T1.invoice_number ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Invoices (invoice_number VARCHAR(255), invoice_date VARCHAR(255)); CREATE TABLE Financial_transactions (invoice_number VARCHAR(255), transaction_id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 909 }
How many invoices do we have?
SELECT COUNT(*) FROM Invoices
CREATE TABLE Invoices (Id VARCHAR(255), invoice_date VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 910 }
Show invoice dates and order id and details for all invoices.
SELECT T1.invoice_date, T1.order_id, T2.order_details FROM Invoices AS T1 JOIN Orders AS T2 ON T1.order_id = T2.order_id
CREATE TABLE Invoices (invoice_date VARCHAR(255), order_id VARCHAR(255)); CREATE TABLE Orders (order_details VARCHAR(255), order_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 911 }
Show the order ids and the number of invoices for each order.
SELECT order_id, COUNT(*) FROM Invoices GROUP BY order_id
CREATE TABLE Invoices (order_id VARCHAR(255), invoice_number VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 912 }
What is the order id and order details for the order more than two invoices.
SELECT T2.order_id, T2.order_details FROM Invoices AS T1 JOIN Orders AS T2 ON T1.order_id = T2.order_id GROUP BY T2.order_id HAVING COUNT(*) > 2
CREATE TABLE Orders (order_id VARCHAR(255), order_details VARCHAR(255)); CREATE TABLE Invoices (order_id VARCHAR(255), invoice_number VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 913 }
What is the customer last name, id and phone number with most number of orders?
SELECT T2.customer_last_name, T1.customer_id, T2.phone_number FROM Orders AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Orders (customer_id VARCHAR(255), order_id VARCHAR(255)); CREATE TABLE Customers (customer_last_name VARCHAR(255), phone_number VARCHAR(255), customer_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 914 }
Show all product names and the total quantity ordered for each product name.
SELECT T2.product_name, SUM(T1.product_quantity) FROM Order_items AS T1 JOIN Products AS T2 ON T1.product_id = T2.product_id GROUP BY T2.product_name
CREATE TABLE Products (product_name VARCHAR(255), product_id VARCHAR(255)); CREATE TABLE Order_items (product_quantity INTEGER, product_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 916 }
Show the order ids and the number of items in each order.
SELECT order_id, COUNT(*) FROM Order_items GROUP BY order_id
CREATE TABLE Order_items (order_id VARCHAR(255), item_id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 917 }
Show the product ids and the number of unique orders containing each product.
SELECT product_id, COUNT(DISTINCT order_id) FROM Order_items GROUP BY product_id
CREATE TABLE Order_items (product_id VARCHAR(255), order_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 918 }
Show all product names and the number of customers having an order on each product.
SELECT T2.product_name, COUNT(*) FROM Order_items AS T1 JOIN Products AS T2 ON T1.product_id = T2.product_id JOIN Orders AS T3 ON T3.order_id = T1.order_id GROUP BY T2.product_name
CREATE TABLE Order_items (product_id VARCHAR(255), order_id VARCHAR(255)); CREATE TABLE Orders (order_id VARCHAR(255), order_date VARCHAR(255)); CREATE TABLE Products (product_name VARCHAR(255), product_id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 919 }
Show order ids and the number of products in each order.
SELECT order_id, COUNT(DISTINCT product_id) FROM Order_items GROUP BY order_id
CREATE TABLE Order_items (order_id VARCHAR(255), product_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 920 }
Show order ids and the total quantity in each order.
SELECT order_id, SUM(product_quantity) FROM Order_items GROUP BY order_id
CREATE TABLE Order_items (order_id VARCHAR(255), product_quantity INTEGER)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 921 }
How many products were not included in any order?
SELECT COUNT(*) FROM products WHERE NOT product_id IN (SELECT product_id FROM Order_items)
CREATE TABLE products (product_id VARCHAR(255), product_name VARCHAR(255)); CREATE TABLE Order_items (product_id VARCHAR(255), quantity INTEGER)
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
1
{ "split": "train", "row_index": 922 }
How many churches opened before 1850 are there?
SELECT COUNT(*) FROM Church WHERE Open_Date < 1850
CREATE TABLE Church (Open_Date INTEGER, church_id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 923 }
Show the name, open date, and organizer for all churches.
SELECT name, open_date, organized_by FROM Church
CREATE TABLE Church (name VARCHAR(255), open_date VARCHAR(255), organized_by VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 924 }
List all church names in descending order of opening date.
SELECT name FROM church ORDER BY open_date DESC
CREATE TABLE church (name VARCHAR(255), open_date VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 925 }
Show the opening year in whcih at least two churches opened.
SELECT open_date FROM church GROUP BY open_date HAVING COUNT(*) >= 2
CREATE TABLE church (open_date 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": 926 }
Show the organizer and name for churches that opened between 1830 and 1840.
SELECT organized_by, name FROM church WHERE open_date BETWEEN 1830 AND 1840
CREATE TABLE church (organized_by VARCHAR(255), name VARCHAR(255), open_date INTEGER)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 927 }
Show all opening years and the number of churches that opened in that year.
SELECT open_date, COUNT(*) FROM church GROUP BY open_date
CREATE TABLE church (open_date 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": 928 }
Show the name and opening year for three churches that opened most recently.
SELECT name, open_date FROM church ORDER BY open_date DESC LIMIT 3
CREATE TABLE church (name VARCHAR(255), open_date VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 929 }
How many female people are older than 30 in our record?
SELECT COUNT(*) FROM people WHERE is_male = 'F' AND age > 30
CREATE TABLE people (is_male VARCHAR(255), age VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 930 }
Show the country where people older than 30 and younger than 25 are from.
SELECT country FROM people WHERE age < 25 INTERSECT SELECT country FROM people WHERE age > 30
CREATE TABLE people (country VARCHAR(255), age INTEGER)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 931 }
Show the minimum, maximum, and average age for all people.
SELECT MIN(age), MAX(age), AVG(age) FROM people
CREATE TABLE people (age 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": 932 }
Show the name and country for all people whose age is smaller than the average.
SELECT name, country FROM people WHERE age < (SELECT AVG(age) FROM people)
CREATE TABLE people (name VARCHAR(255), country VARCHAR(255), age INTEGER)
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
1
{ "split": "train", "row_index": 933 }
Show the pair of male and female names in all weddings after year 2014
SELECT T2.name, T3.name FROM wedding AS T1 JOIN people AS T2 ON T1.male_id = T2.people_id JOIN people AS T3 ON T1.female_id = T3.people_id WHERE T1.year > 2014
CREATE TABLE wedding (male_id VARCHAR(255), female_id VARCHAR(255), year INTEGER); CREATE TABLE people (name VARCHAR(255), people_id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 934 }
Show the name and age for all male people who don't have a wedding.
SELECT name, age FROM people WHERE is_male = 'T' AND NOT people_id IN (SELECT male_id FROM wedding)
CREATE TABLE wedding (name VARCHAR(255), age VARCHAR(255), is_male VARCHAR(255), people_id VARCHAR(255), male_id VARCHAR(255)); CREATE TABLE people (name VARCHAR(255), age VARCHAR(255), is_male VARCHAR(255), people_id VARCHAR(255), male_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
2
{ "split": "train", "row_index": 935 }
Show all church names that have hosted least two weddings.
SELECT T1.name FROM church AS T1 JOIN wedding AS T2 ON T1.church_id = T2.church_id GROUP BY T1.church_id HAVING COUNT(*) >= 2
CREATE TABLE wedding (church_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE church (name VARCHAR(255), church_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 937 }
Show the names for all females from Canada having a wedding in year 2016.
SELECT T2.name FROM wedding AS T1 JOIN people AS T2 ON T1.female_id = T2.people_id WHERE T1.year = 2016 AND T2.is_male = 'F' AND T2.country = 'Canada'
CREATE TABLE people (name VARCHAR(255), people_id VARCHAR(255), country VARCHAR(255), is_male VARCHAR(255)); CREATE TABLE wedding (female_id VARCHAR(255), year VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 938 }
How many weddings are there in year 2016?
SELECT COUNT(*) FROM wedding WHERE YEAR = 2016
CREATE TABLE wedding (YEAR 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": 939 }
Show the church names for the weddings of all people older than 30.
SELECT T4.name FROM wedding AS T1 JOIN people AS T2 ON T1.male_id = T2.people_id JOIN people AS T3 ON T1.female_id = T3.people_id JOIN church AS T4 ON T4.church_id = T1.church_id WHERE T2.age > 30 OR T3.age > 30
CREATE TABLE church (name VARCHAR(255), church_id VARCHAR(255)); CREATE TABLE people (people_id VARCHAR(255), age VARCHAR(255)); CREATE TABLE wedding (male_id VARCHAR(255), female_id VARCHAR(255), church_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
5
0
1
{ "split": "train", "row_index": 940 }
Show all countries and the number of people from each country.
SELECT country, COUNT(*) FROM people GROUP BY country
CREATE TABLE people (country 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": 941 }
How many churches have a wedding in year 2016?
SELECT COUNT(DISTINCT church_id) FROM wedding WHERE YEAR = 2016
CREATE TABLE wedding (church_id VARCHAR(255), YEAR VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 942 }
How many artists do we have?
SELECT COUNT(*) FROM artist
CREATE TABLE artist (Id VARCHAR(255), name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 943 }
Show all artist name, age, and country ordered by the yeared they joined.
SELECT name, age, country FROM artist ORDER BY Year_Join
CREATE TABLE artist (name VARCHAR(255), age VARCHAR(255), country VARCHAR(255), Year_Join VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 944 }
What are all distinct country for artists?
SELECT DISTINCT country FROM artist
CREATE TABLE artist (country 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": 945 }
Show all artist names and the year joined who are not from United States.
SELECT name, year_join FROM artist WHERE country <> 'United States'
CREATE TABLE artist (name VARCHAR(255), year_join VARCHAR(255), country VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 946 }
How many artists are above age 46 and joined after 1990?
SELECT COUNT(*) FROM artist WHERE age > 46 AND year_join > 1990
CREATE TABLE artist (age VARCHAR(255), year_join VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 947 }
What is the average and minimum age of all artists from United States.
SELECT AVG(age), MIN(age) FROM artist WHERE country = 'United States'
CREATE TABLE artist (age INTEGER, country VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
2
{ "split": "train", "row_index": 948 }
What is the name of the artist who joined latest?
SELECT name FROM artist ORDER BY year_join DESC LIMIT 1
CREATE TABLE artist (name VARCHAR(255), year_join VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 949 }
How many exhibition are there in year 2005 or after?
SELECT COUNT(*) FROM exhibition WHERE YEAR >= 2005
CREATE TABLE exhibition (YEAR 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": 950 }
Show theme and year for all exhibitions with ticket prices lower than 15.
SELECT theme, YEAR FROM exhibition WHERE ticket_price < 15
CREATE TABLE exhibition (theme VARCHAR(255), YEAR VARCHAR(255), ticket_price INTEGER)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 951 }
Show all artist names and the number of exhibitions for each artist.
SELECT T2.name, COUNT(*) FROM exhibition AS T1 JOIN artist AS T2 ON T1.artist_id = T2.artist_id GROUP BY T1.artist_id
CREATE TABLE artist (name VARCHAR(255), artist_id VARCHAR(255)); CREATE TABLE exhibition (artist_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": 952 }
What is the name and country for the artist with most number of exhibitions?
SELECT T2.name, T2.country FROM exhibition AS T1 JOIN artist AS T2 ON T1.artist_id = T2.artist_id GROUP BY T1.artist_id ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE exhibition (artist_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE artist (name VARCHAR(255), country VARCHAR(255), artist_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 953 }
Show names for artists without any exhibition.
SELECT name FROM artist WHERE NOT artist_id IN (SELECT artist_id FROM exhibition)
CREATE TABLE artist (name VARCHAR(255), artist_id VARCHAR(255)); CREATE TABLE exhibition (name VARCHAR(255), artist_id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 954 }
Show theme and year for all exhibitions in an descending order of ticket price.
SELECT theme, YEAR FROM exhibition ORDER BY ticket_price DESC
CREATE TABLE exhibition (theme VARCHAR(255), YEAR VARCHAR(255), ticket_price VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 957 }
What is the theme, date, and attendance for the exhibition in year 2004?
SELECT T2.theme, T1.date, T1.attendance FROM exhibition_record AS T1 JOIN exhibition AS T2 ON T1.exhibition_id = T2.exhibition_id WHERE T2.year = 2004
CREATE TABLE exhibition_record (date VARCHAR(255), attendance VARCHAR(255), exhibition_id VARCHAR(255)); CREATE TABLE exhibition (theme VARCHAR(255), exhibition_id VARCHAR(255), year VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 958 }
Show the theme for exhibitions with both records of an attendance below 100 and above 500.
SELECT T2.theme FROM exhibition_record AS T1 JOIN exhibition AS T2 ON T1.exhibition_id = T2.exhibition_id WHERE T1.attendance < 100 INTERSECT SELECT T2.theme FROM exhibition_record AS T1 JOIN exhibition AS T2 ON T1.exhibition_id = T2.exhibition_id WHERE T1.attendance > 500
CREATE TABLE exhibition (theme VARCHAR(255), exhibition_id VARCHAR(255)); CREATE TABLE exhibition_record (exhibition_id VARCHAR(255), attendance INTEGER)
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
1
0
{ "split": "train", "row_index": 960 }
How many exhibitions have a attendance more than 100 or have a ticket price below 10?
SELECT COUNT(*) FROM exhibition_record AS T1 JOIN exhibition AS T2 ON T1.exhibition_id = T2.exhibition_id WHERE T1.attendance > 100 OR T2.ticket_price < 10
CREATE TABLE exhibition_record (exhibition_id VARCHAR(255), attendance VARCHAR(255)); CREATE TABLE exhibition (exhibition_id VARCHAR(255), ticket_price VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 961 }
Show all artist names with an average exhibition attendance over 200.
SELECT T3.name FROM exhibition_record AS T1 JOIN exhibition AS T2 ON T1.exhibition_id = T2.exhibition_id JOIN artist AS T3 ON T3.artist_id = T2.artist_id GROUP BY T3.artist_id HAVING AVG(T1.attendance) > 200
CREATE TABLE artist (name VARCHAR(255), artist_id VARCHAR(255)); CREATE TABLE exhibition (exhibition_id VARCHAR(255), artist_id VARCHAR(255)); CREATE TABLE exhibition_record (exhibition_id VARCHAR(255), attendance INTEGER)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 962 }
Find the id of the item whose title is "orange".
SELECT i_id FROM item WHERE title = 'orange'
CREATE TABLE item (i_id VARCHAR(255), title VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 963 }
Find the number of reviews.
SELECT COUNT(*) FROM review
CREATE TABLE review (Id VARCHAR(255), review_text VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 965 }
How many users are there?
SELECT COUNT(*) FROM useracct
CREATE TABLE useracct (Id VARCHAR(255), username VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 966 }
Find the average and maximum rating of all reviews.
SELECT AVG(rating), MAX(rating) FROM review
CREATE TABLE review (rating 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": 967 }
Find the highest rank of all reviews.
SELECT MIN(rank) FROM review
CREATE TABLE review (rank INTEGER, id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 968 }
How many different users wrote some reviews?
SELECT COUNT(DISTINCT u_id) FROM review
CREATE TABLE review (u_id VARCHAR(255), rating INTEGER)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 969 }
How many different items were reviewed by some users?
SELECT COUNT(DISTINCT i_id) FROM review
CREATE TABLE review (i_id VARCHAR(255), rating INTEGER)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 970 }
Find the number of items that did not receive any review.
SELECT COUNT(*) FROM item WHERE NOT i_id IN (SELECT i_id FROM review)
CREATE TABLE review (i_id VARCHAR(255), rating INTEGER); CREATE TABLE item (i_id VARCHAR(255), title VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
1
{ "split": "train", "row_index": 971 }
Find the names of users who did not leave any review.
SELECT name FROM useracct WHERE NOT u_id IN (SELECT u_id FROM review)
CREATE TABLE review (name VARCHAR(255), u_id VARCHAR(255)); CREATE TABLE useracct (name VARCHAR(255), u_id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 972 }
Find the names of goods that receive a rating of 10.
SELECT T1.title FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id WHERE T2.rating = 10
CREATE TABLE item (title VARCHAR(255), i_id VARCHAR(255)); CREATE TABLE review (i_id VARCHAR(255), rating VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 973 }
Find the titles of items that received any rating below 5.
SELECT T1.title FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id WHERE T2.rating < 5
CREATE TABLE item (title VARCHAR(255), i_id VARCHAR(255)); CREATE TABLE review (i_id VARCHAR(255), rating INTEGER)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 975 }
Find the titles of items that received both a rating higher than 8 and a rating below 5.
SELECT T1.title FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id WHERE T2.rating > 8 INTERSECT SELECT T1.title FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id WHERE T2.rating < 5
CREATE TABLE item (title VARCHAR(255), i_id VARCHAR(255)); CREATE TABLE review (i_id VARCHAR(255), rating INTEGER)
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
1
0
{ "split": "train", "row_index": 976 }
Find the names of items whose rank is higher than 3 and whose average rating is above 5.
SELECT T1.title FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id WHERE T2.rank > 3 INTERSECT SELECT T1.title FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id GROUP BY T2.i_id HAVING AVG(T2.rating) > 5
CREATE TABLE item (title VARCHAR(255), i_id VARCHAR(255)); CREATE TABLE review (i_id VARCHAR(255), rank INTEGER, rating INTEGER)
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
1
0
{ "split": "train", "row_index": 977 }
Find the name of the item with the lowest average rating.
SELECT T1.title FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id GROUP BY T2.i_id ORDER BY AVG(T2.rating) LIMIT 1
CREATE TABLE item (title VARCHAR(255), i_id VARCHAR(255)); CREATE TABLE review (i_id VARCHAR(255), rating INTEGER)
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
0
{ "split": "train", "row_index": 978 }
List the titles of all items in alphabetic order .
SELECT title FROM item ORDER BY title
CREATE TABLE item (title 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": 979 }
Find the name of the user who gives the most reviews.
SELECT T1.name FROM useracct AS T1 JOIN review AS T2 ON T1.u_id = T2.u_id GROUP BY T2.u_id ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE useracct (name VARCHAR(255), u_id VARCHAR(255)); CREATE TABLE review (u_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
0
{ "split": "train", "row_index": 980 }
Find the name and id of the item with the highest average rating.
SELECT T1.title, T1.i_id FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id GROUP BY T2.i_id ORDER BY AVG(T2.rating) DESC LIMIT 1
CREATE TABLE item (title VARCHAR(255), i_id VARCHAR(255)); CREATE TABLE review (i_id VARCHAR(255), rating INTEGER)
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 981 }
Find the name and id of the good with the highest average rank.
SELECT T1.title, T1.i_id FROM item AS T1 JOIN review AS T2 ON T1.i_id = T2.i_id GROUP BY T2.i_id ORDER BY AVG(T2.rank) DESC LIMIT 1
CREATE TABLE review (i_id VARCHAR(255), rank INTEGER); CREATE TABLE item (title VARCHAR(255), i_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 982 }
For each user, return the name and the average rating of reviews given by them.
SELECT T1.name, AVG(T2.rating) FROM useracct AS T1 JOIN review AS T2 ON T1.u_id = T2.u_id GROUP BY T2.u_id
CREATE TABLE review (rating INTEGER, u_id VARCHAR(255)); CREATE TABLE useracct (name VARCHAR(255), u_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 983 }
For each user, find their name and the number of reviews written by them.
SELECT T1.name, COUNT(*) FROM useracct AS T1 JOIN review AS T2 ON T1.u_id = T2.u_id GROUP BY T2.u_id
CREATE TABLE useracct (name VARCHAR(255), u_id VARCHAR(255)); CREATE TABLE review (u_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 984 }
Find the name of the user who gave the highest rating.
SELECT T1.name FROM useracct AS T1 JOIN review AS T2 ON T1.u_id = T2.u_id ORDER BY T2.rating DESC LIMIT 1
CREATE TABLE review (u_id VARCHAR(255), rating VARCHAR(255)); CREATE TABLE useracct (name VARCHAR(255), u_id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 985 }
Find the names of the items that did not receive any review.
SELECT title FROM item WHERE NOT i_id IN (SELECT i_id FROM review)
CREATE TABLE item (title VARCHAR(255), i_id VARCHAR(255)); CREATE TABLE review (title VARCHAR(255), i_id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 989 }
Find the number of users who did not write any review.
SELECT COUNT(*) FROM useracct WHERE NOT u_id IN (SELECT u_id FROM review)
CREATE TABLE review (u_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE useracct (u_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
1
{ "split": "train", "row_index": 990 }
How many players are there?
SELECT COUNT(*) FROM player
CREATE TABLE player (Id VARCHAR(255), name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 991 }
List the names of players in ascending order of votes.
SELECT Player_name FROM player ORDER BY Votes
CREATE TABLE player (Player_name VARCHAR(255), Votes VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 992 }
What are the gender and occupation of players?
SELECT Gender, Occupation FROM player
CREATE TABLE player (Gender VARCHAR(255), Occupation VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 993 }