question_id
int64
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33 values
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4 values
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3 values
SQL
stringlengths
25
862
1,501
restaurant_1
spider
What is the age of student Linda Smith?
train
medium
SELECT age FROM student WHERE fname = 'linda' AND lname = 'smith'
1,502
restaurant_1
spider
What is the gender of the student Linda Smith?
train
medium
SELECT sex FROM student WHERE fname = 'linda' AND lname = 'smith'
1,503
restaurant_1
spider
List all students' first names and last names who majored in 600.
train
medium
SELECT fname, lname FROM student WHERE major = 600
1,504
restaurant_1
spider
Which city does student Linda Smith live in?
train
medium
SELECT city_code FROM student WHERE fname = 'linda' AND lname = 'smith'
1,505
restaurant_1
spider
Advisor 1121 has how many students?
train
medium
SELECT COUNT(*) FROM student WHERE advisor = 1121
1,506
restaurant_1
spider
Which Advisor has most of students? List advisor and the number of students.
train
hard
SELECT advisor, COUNT(*) FROM student GROUP BY advisor ORDER BY COUNT(advisor) DESC LIMIT 1
1,507
restaurant_1
spider
Which major has least number of students? List the major and the number of students.
train
hard
SELECT major, COUNT(*) FROM student GROUP BY major ORDER BY COUNT(major) ASC LIMIT 1
1,508
restaurant_1
spider
Which major has between 2 and 30 number of students? List major and the number of students.
train
hard
SELECT major, COUNT(*) FROM student GROUP BY major HAVING COUNT(major) BETWEEN 2 AND 30
1,509
restaurant_1
spider
Which student's age is older than 18 and is majoring in 600? List each student's first and last name.
train
medium
SELECT fname, lname FROM student WHERE age > 18 AND major = 600
1,510
restaurant_1
spider
List all female students age is older than 18 who is not majoring in 600. List students' first name and last name.
train
medium
SELECT fname, lname FROM student WHERE age > 18 AND major <> 600 AND sex = 'f'
1,511
restaurant_1
spider
How many restaurant is the Sandwich type restaurant?
train
hard
SELECT COUNT(*) FROM restaurant JOIN type_of_restaurant ON restaurant.resid = type_of_restaurant.resid JOIN restaurant_type ON type_of_restaurant.restypeid = restaurant_type.restypeid GROUP BY type_of_restaurant.restypeid HAVING restaurant_type.restypename = 'sandwich'
1,512
restaurant_1
spider
How long does student Linda Smith spend on the restaurant in total?
train
hard
SELECT SUM(spent) FROM student JOIN visits_restaurant ON student.stuid = visits_restaurant.stuid WHERE student.fname = 'linda' AND student.lname = 'smith'
1,513
restaurant_1
spider
How many times has the student Linda Smith visited Subway?
train
hard
SELECT COUNT(*) FROM student JOIN visits_restaurant ON student.stuid = visits_restaurant.stuid JOIN restaurant ON visits_restaurant.resid = restaurant.resid WHERE student.fname = 'linda' AND student.lname = 'smith' AND restaurant.resname = 'subway'
1,514
restaurant_1
spider
When did Linda Smith visit Subway?
train
hard
SELECT time FROM student JOIN visits_restaurant ON student.stuid = visits_restaurant.stuid JOIN restaurant ON visits_restaurant.resid = restaurant.resid WHERE student.fname = 'linda' AND student.lname = 'smith' AND restaurant.resname = 'subway'
1,515
restaurant_1
spider
At which restaurant did the students spend the least amount of time? List restaurant and the time students spent on in total.
train
hard
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) ASC LIMIT 1
1,516
restaurant_1
spider
Which student visited restaurant most often? List student's first name and last name.
train
hard
SELECT student.fname, student.lname FROM student JOIN visits_restaurant ON student.stuid = visits_restaurant.stuid GROUP BY student.stuid ORDER BY COUNT(*) DESC LIMIT 1
1,517
customer_deliveries
spider
Find the ids of orders whose status is 'Success'.
train
medium
SELECT actual_order_id FROM actual_orders WHERE order_status_code = 'success'
1,518
customer_deliveries
spider
Find the name and price of the product that has been ordered the greatest number of times.
train
hard
SELECT t1.product_name, t1.product_price FROM products AS t1 JOIN regular_order_products AS t2 ON t1.product_id = t2.product_id GROUP BY t2.product_id ORDER BY COUNT(*) DESC LIMIT 1
1,519
customer_deliveries
spider
Find the number of customers in total.
train
easy
SELECT COUNT(*) FROM customers
1,520
customer_deliveries
spider
How many different payment methods are there?
train
easy
SELECT COUNT(DISTINCT payment_method) FROM customers
1,521
customer_deliveries
spider
Show the details of all trucks in the order of their license number.
train
hard
SELECT truck_details FROM trucks ORDER BY truck_licence_number
1,522
customer_deliveries
spider
Find the name of the most expensive product.
train
hard
SELECT product_name FROM products ORDER BY product_price DESC LIMIT 1
1,523
customer_deliveries
spider
Find the names of customers who are not living in the state of California.
train
hard
SELECT customer_name FROM customers EXCEPT SELECT t1.customer_name 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 WHERE t3.state_province_county = 'california'
1,524
customer_deliveries
spider
List the names and emails of customers who payed by Visa card.
train
easy
SELECT customer_email, customer_name FROM customers WHERE payment_method = 'visa'
1,525
customer_deliveries
spider
Find the names and phone numbers of customers living in California state.
train
hard
SELECT t1.customer_name, t1.customer_phone 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 WHERE t3.state_province_county = 'california'
1,526
customer_deliveries
spider
Find the states which do not have any employee in their record.
train
easy
SELECT state_province_county FROM addresses WHERE NOT address_id IN (SELECT employee_address_id FROM employees)
1,527
customer_deliveries
spider
List the names, phone numbers, and emails of all customers sorted by their dates of becoming customers.
train
hard
SELECT customer_name, customer_phone, customer_email FROM customers ORDER BY date_became_customer
1,528
customer_deliveries
spider
Find the name of the first 5 customers.
train
hard
SELECT customer_name FROM customers ORDER BY date_became_customer LIMIT 5
1,529
customer_deliveries
spider
Find the payment method that is used most frequently.
train
hard
SELECT payment_method FROM customers GROUP BY payment_method ORDER BY COUNT(*) DESC LIMIT 1
1,530
customer_deliveries
spider
List the names of all routes in alphabetic order.
train
hard
SELECT route_name FROM delivery_routes ORDER BY route_name
1,531
customer_deliveries
spider
Find the name of route that has the highest number of deliveries.
train
hard
SELECT t1.route_name FROM delivery_routes AS t1 JOIN delivery_route_locations AS t2 ON t1.route_id = t2.route_id GROUP BY t1.route_id ORDER BY COUNT(*) DESC LIMIT 1
1,532
customer_deliveries
spider
List the state names and the number of customers living in each state.
train
hard
SELECT t2.state_province_county, COUNT(*) FROM customer_addresses AS t1 JOIN addresses AS t2 ON t1.address_id = t2.address_id GROUP BY t2.state_province_county
1,533
loan_1
spider
How many bank branches are there?
train
easy
SELECT COUNT(*) FROM bank
1,534
loan_1
spider
Count the number of bank branches.
train
easy
SELECT COUNT(*) FROM bank
1,535
loan_1
spider
How many customers are there?
train
easy
SELECT SUM(no_of_customers) FROM bank
1,536
loan_1
spider
What is the total number of customers across banks?
train
easy
SELECT SUM(no_of_customers) FROM bank
1,537
loan_1
spider
Find the number of customers in the banks at New York City.
train
medium
SELECT SUM(no_of_customers) FROM bank WHERE city = 'new york city'
1,538
loan_1
spider
What is the total number of customers who use banks in New York City?
train
medium
SELECT SUM(no_of_customers) FROM bank WHERE city = 'new york city'
1,539
loan_1
spider
Find the average number of customers in all banks of Utah state.
train
easy
SELECT AVG(no_of_customers) FROM bank WHERE state = 'utah'
1,540
loan_1
spider
What is the average number of customers across banks in the state of Utah?
train
easy
SELECT AVG(no_of_customers) FROM bank WHERE state = 'utah'
1,541
loan_1
spider
Find the average number of customers cross all banks.
train
easy
SELECT AVG(no_of_customers) FROM bank
1,542
loan_1
spider
What is the average number of bank customers?
train
easy
SELECT AVG(no_of_customers) FROM bank
1,543
loan_1
spider
Find the city and state of the bank branch named morningside.
train
medium
SELECT city, state FROM bank WHERE bname = 'morningside'
1,544
loan_1
spider
What city and state is the bank with the name morningside in?
train
medium
SELECT city, state FROM bank WHERE bname = 'morningside'
1,545
loan_1
spider
Find the branch names of banks in the New York state.
train
medium
SELECT bname FROM bank WHERE state = 'new york'
1,546
loan_1
spider
What are the names of banks in the state of New York?
train
medium
SELECT bname FROM bank WHERE state = 'new york'
1,547
loan_1
spider
List the name of all customers sorted by their account balance in ascending order.
train
hard
SELECT cust_name FROM customer ORDER BY acc_bal
1,548
loan_1
spider
What are the names of all customers, ordered by account balance?
train
hard
SELECT cust_name FROM customer ORDER BY acc_bal
1,549
loan_1
spider
List the name of all different customers who have some loan sorted by their total loan amount.
train
hard
SELECT t1.cust_name FROM customer AS t1 JOIN loan AS t2 ON t1.cust_id = t2.cust_id GROUP BY t1.cust_name ORDER BY SUM(t2.amount)
1,550
loan_1
spider
What are the names of the different customers who have taken out a loan, ordered by the total amount that they have taken?
train
hard
SELECT t1.cust_name FROM customer AS t1 JOIN loan AS t2 ON t1.cust_id = t2.cust_id GROUP BY t1.cust_name ORDER BY SUM(t2.amount)
1,551
loan_1
spider
Find the state, account type, and credit score of the customer whose number of loan is 0.
train
easy
SELECT state, acc_type, credit_score FROM customer WHERE no_of_loans = 0
1,552
loan_1
spider
What are the states, account types, and credit scores for customers who have 0 loans?
train
easy
SELECT state, acc_type, credit_score FROM customer WHERE no_of_loans = 0
1,553
loan_1
spider
Find the number of different cities which banks are located at.
train
easy
SELECT COUNT(DISTINCT city) FROM bank
1,554
loan_1
spider
In how many different cities are banks located?
train
easy
SELECT COUNT(DISTINCT city) FROM bank
1,555
loan_1
spider
Find the number of different states which banks are located at.
train
easy
SELECT COUNT(DISTINCT state) FROM bank
1,556
loan_1
spider
In how many different states are banks located?
train
easy
SELECT COUNT(DISTINCT state) FROM bank
1,557
loan_1
spider
How many distinct types of accounts are there?
train
easy
SELECT COUNT(DISTINCT acc_type) FROM customer
1,558
loan_1
spider
Count the number of different account types.
train
easy
SELECT COUNT(DISTINCT acc_type) FROM customer
1,559
loan_1
spider
Find the name and account balance of the customer whose name includes the letter ‘a’.
train
easy
SELECT cust_name, acc_bal FROM customer WHERE cust_name LIKE '%a%'
1,560
loan_1
spider
What are the names and account balances of customers with the letter a in their names?
train
easy
SELECT cust_name, acc_bal FROM customer WHERE cust_name LIKE '%a%'
1,561
loan_1
spider
Find the total account balance of each customer from Utah or Texas.
train
medium
SELECT SUM(acc_bal) FROM customer WHERE state = 'utah' OR state = 'texas'
1,562
loan_1
spider
What are the total account balances for each customer from Utah or Texas?
train
medium
SELECT SUM(acc_bal) FROM customer WHERE state = 'utah' OR state = 'texas'
1,563
loan_1
spider
Find the name of customers who have both saving and checking account types.
train
easy
SELECT cust_name FROM customer WHERE acc_type = 'saving' INTERSECT SELECT cust_name FROM customer WHERE acc_type = 'checking'
1,564
loan_1
spider
What are the names of customers who have both savings and checking accounts?
train
easy
SELECT cust_name FROM customer WHERE acc_type = 'saving' INTERSECT SELECT cust_name FROM customer WHERE acc_type = 'checking'
1,565
loan_1
spider
Find the name of customers who do not have an saving account.
train
easy
SELECT cust_name FROM customer EXCEPT SELECT cust_name FROM customer WHERE acc_type = 'saving'
1,566
loan_1
spider
What are the names of customers who do not have saving accounts?
train
easy
SELECT cust_name FROM customer EXCEPT SELECT cust_name FROM customer WHERE acc_type = 'saving'
1,567
loan_1
spider
Find the name of customers who do not have a loan with a type of Mortgages.
train
hard
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'
1,568
loan_1
spider
What are the names of customers who have not taken a Mortage loan?
train
hard
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'
1,569
loan_1
spider
Find the name of customers who have loans of both Mortgages and Auto.
train
hard
SELECT t1.cust_name FROM customer AS t1 JOIN loan AS t2 ON t1.cust_id = t2.cust_id WHERE loan_type = 'mortgages' INTERSECT SELECT t1.cust_name FROM customer AS t1 JOIN loan AS t2 ON t1.cust_id = t2.cust_id WHERE loan_type = 'auto'
1,570
loan_1
spider
What are the names of customers who have taken both Mortgage and Auto loans?
train
hard
SELECT t1.cust_name FROM customer AS t1 JOIN loan AS t2 ON t1.cust_id = t2.cust_id WHERE loan_type = 'mortgages' INTERSECT SELECT t1.cust_name FROM customer AS t1 JOIN loan AS t2 ON t1.cust_id = t2.cust_id WHERE loan_type = 'auto'
1,571
loan_1
spider
Find the name of customers whose credit score is below the average credit scores of all customers.
train
medium
SELECT cust_name FROM customer WHERE credit_score < (SELECT AVG(credit_score) FROM customer)
1,572
loan_1
spider
What are the names of customers with credit score less than the average credit score across customers?
train
medium
SELECT cust_name FROM customer WHERE credit_score < (SELECT AVG(credit_score) FROM customer)
1,573
loan_1
spider
Find the branch name of the bank that has the most number of customers.
train
hard
SELECT bname FROM bank ORDER BY no_of_customers DESC LIMIT 1
1,574
loan_1
spider
What is the name of the bank branch with the greatest number of customers?
train
hard
SELECT bname FROM bank ORDER BY no_of_customers DESC LIMIT 1
1,575
loan_1
spider
Find the name of customer who has the lowest credit score.
train
hard
SELECT cust_name FROM customer ORDER BY credit_score LIMIT 1
1,576
loan_1
spider
What is the name of the customer with the worst credit score?
train
hard
SELECT cust_name FROM customer ORDER BY credit_score LIMIT 1
1,577
loan_1
spider
Find the name, account type, and account balance of the customer who has the highest credit score.
train
hard
SELECT cust_name, acc_type, acc_bal FROM customer ORDER BY credit_score DESC LIMIT 1
1,578
loan_1
spider
What is the name, account type, and account balance corresponding to the customer with the highest credit score?
train
hard
SELECT cust_name, acc_type, acc_bal FROM customer ORDER BY credit_score DESC LIMIT 1
1,579
loan_1
spider
Find the name of customer who has the highest amount of loans.
train
hard
SELECT t1.cust_name FROM customer AS t1 JOIN loan AS t2 ON t1.cust_id = t2.cust_id GROUP BY t1.cust_name ORDER BY SUM(t2.amount) DESC LIMIT 1
1,580
loan_1
spider
What is the name of the customer who has greatest total loan amount?
train
hard
SELECT t1.cust_name FROM customer AS t1 JOIN loan AS t2 ON t1.cust_id = t2.cust_id GROUP BY t1.cust_name ORDER BY SUM(t2.amount) DESC LIMIT 1
1,581
loan_1
spider
Find the state which has the most number of customers.
train
hard
SELECT state FROM bank GROUP BY state ORDER BY SUM(no_of_customers) DESC LIMIT 1
1,582
loan_1
spider
Which state has the greatest total number of bank customers?
train
hard
SELECT state FROM bank GROUP BY state ORDER BY SUM(no_of_customers) DESC LIMIT 1
1,583
loan_1
spider
For each account type, find the average account balance of customers with credit score lower than 50.
train
medium
SELECT AVG(acc_bal), acc_type FROM customer WHERE credit_score < 50 GROUP BY acc_type
1,584
loan_1
spider
What is the average account balance of customers with credit score below 50 for the different account types?
train
medium
SELECT AVG(acc_bal), acc_type FROM customer WHERE credit_score < 50 GROUP BY acc_type
1,585
loan_1
spider
For each state, find the total account balance of customers whose credit score is above 100.
train
medium
SELECT SUM(acc_bal), state FROM customer WHERE credit_score > 100 GROUP BY state
1,586
loan_1
spider
What is the total account balance for customers with a credit score of above 100 for the different states?
train
medium
SELECT SUM(acc_bal), state FROM customer WHERE credit_score > 100 GROUP BY state
1,587
loan_1
spider
Find the total amount of loans offered by each bank branch.
train
hard
SELECT SUM(amount), t1.bname FROM bank AS t1 JOIN loan AS t2 ON t1.branch_id = t2.branch_id GROUP BY t1.bname
1,588
loan_1
spider
What are the names of the different bank branches, and what are their total loan amounts?
train
hard
SELECT SUM(amount), t1.bname FROM bank AS t1 JOIN loan AS t2 ON t1.branch_id = t2.branch_id GROUP BY t1.bname
1,589
loan_1
spider
Find the name of customers who have more than one loan.
train
hard
SELECT t1.cust_name FROM customer AS t1 JOIN loan AS t2 ON t1.cust_id = t2.cust_id GROUP BY t1.cust_name HAVING COUNT(*) > 1
1,590
loan_1
spider
What are the names of customers who have taken out more than one loan?
train
hard
SELECT t1.cust_name FROM customer AS t1 JOIN loan AS t2 ON t1.cust_id = t2.cust_id GROUP BY t1.cust_name HAVING COUNT(*) > 1
1,591
loan_1
spider
Find the name and account balance of the customers who have loans with a total amount of more than 5000.
train
hard
SELECT t1.cust_name, t1.acc_type FROM customer AS t1 JOIN loan AS t2 ON t1.cust_id = t2.cust_id GROUP BY t1.cust_name HAVING SUM(t2.amount) > 5000
1,592
loan_1
spider
What are the names and account balances for customers who have taken a total amount of more than 5000 in loans?
train
hard
SELECT t1.cust_name, t1.acc_type FROM customer AS t1 JOIN loan AS t2 ON t1.cust_id = t2.cust_id GROUP BY t1.cust_name HAVING SUM(t2.amount) > 5000
1,593
loan_1
spider
Find the name of bank branch that provided the greatest total amount of loans.
train
hard
SELECT t1.bname FROM bank AS t1 JOIN loan AS t2 ON t1.branch_id = t2.branch_id GROUP BY t1.bname ORDER BY SUM(t2.amount) DESC LIMIT 1
1,594
loan_1
spider
What is the name of the bank branch that has lent the greatest amount?
train
hard
SELECT t1.bname FROM bank AS t1 JOIN loan AS t2 ON t1.branch_id = t2.branch_id GROUP BY t1.bname ORDER BY SUM(t2.amount) DESC LIMIT 1
1,595
loan_1
spider
Find the name of bank branch that provided the greatest total amount of loans to customers with credit score is less than 100.
train
hard
SELECT t2.bname FROM loan AS t1 JOIN bank AS t2 ON t1.branch_id = t2.branch_id JOIN customer AS t3 ON t1.cust_id = t3.cust_id WHERE t3.credit_score < 100 GROUP BY t2.bname ORDER BY SUM(t1.amount) DESC LIMIT 1
1,596
loan_1
spider
What is the name of the bank branch that has lended the largest total amount in loans, specifically to customers with credit scores below 100?
train
hard
SELECT t2.bname FROM loan AS t1 JOIN bank AS t2 ON t1.branch_id = t2.branch_id JOIN customer AS t3 ON t1.cust_id = t3.cust_id WHERE t3.credit_score < 100 GROUP BY t2.bname ORDER BY SUM(t1.amount) DESC LIMIT 1
1,597
loan_1
spider
Find the name of bank branches that provided some loans.
train
hard
SELECT DISTINCT t1.bname FROM bank AS t1 JOIN loan AS t2 ON t1.branch_id = t2.branch_id
1,598
loan_1
spider
What are the names of the different banks that have provided loans?
train
hard
SELECT DISTINCT t1.bname FROM bank AS t1 JOIN loan AS t2 ON t1.branch_id = t2.branch_id
1,599
loan_1
spider
Find the name and credit score of the customers who have some loans.
train
hard
SELECT DISTINCT t1.cust_name, t1.credit_score FROM customer AS t1 JOIN loan AS t2 ON t1.cust_id = t2.cust_id
1,600
loan_1
spider
What are the different names and credit scores of customers who have taken a loan?
train
hard
SELECT DISTINCT t1.cust_name, t1.credit_score FROM customer AS t1 JOIN loan AS t2 ON t1.cust_id = t2.cust_id