query_id
int64 0
7k
| database_id
stringclasses 140
values | table_id
sequencelengths 1
5
| query
stringlengths 16
224
| answer
stringlengths 18
577
| difficulty
stringclasses 4
values |
|---|---|---|---|---|---|
2,900 | icfp_1 | [
"authorship",
"inst",
"papers"
] | Retrieve the country that has published the most papers. | SELECT t1.country FROM inst AS t1 JOIN authorship AS t2 ON t1.instid = t2.instid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.country ORDER BY count(*) DESC LIMIT 1 | extra |
2,901 | icfp_1 | [
"authorship",
"inst",
"papers"
] | Find the country that the most papers are affiliated with. | SELECT t1.country FROM inst AS t1 JOIN authorship AS t2 ON t1.instid = t2.instid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.country ORDER BY count(*) DESC LIMIT 1 | extra |
2,902 | icfp_1 | [
"authorship",
"inst",
"papers"
] | Find the name of the organization that has published the largest number of papers. | SELECT t1.name FROM inst AS t1 JOIN authorship AS t2 ON t1.instid = t2.instid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.name ORDER BY count(*) DESC LIMIT 1 | extra |
2,903 | icfp_1 | [
"authorship",
"inst",
"papers"
] | Which institution has the most papers? Find the name of the institution. | SELECT t1.name FROM inst AS t1 JOIN authorship AS t2 ON t1.instid = t2.instid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.name ORDER BY count(*) DESC LIMIT 1 | extra |
2,904 | icfp_1 | [
"papers"
] | Find the titles of the papers that contain the word "ML". | SELECT title FROM papers WHERE title LIKE "%ML%" | medium |
2,905 | icfp_1 | [
"papers"
] | Which papers have the substring "ML" in their titles? Return the titles of the papers. | SELECT title FROM papers WHERE title LIKE "%ML%" | medium |
2,906 | icfp_1 | [
"papers"
] | Which paper's title contains the word "Database"? | SELECT title FROM papers WHERE title LIKE "%Database%" | medium |
2,907 | icfp_1 | [
"papers"
] | Which papers have the substring "Database" in their titles? Show the titles of the papers. | SELECT title FROM papers WHERE title LIKE "%Database%" | medium |
2,908 | icfp_1 | [
"authorship",
"authors",
"papers"
] | Find the first names of all the authors who have written a paper with title containing the word "Functional". | SELECT t1.fname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title LIKE "%Functional%" | extra |
2,909 | icfp_1 | [
"authorship",
"authors",
"papers"
] | Who has written a paper that has the word "Functional" in its title? Return the first names of the authors. | SELECT t1.fname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title LIKE "%Functional%" | extra |
2,910 | icfp_1 | [
"authorship",
"authors",
"papers"
] | Find the last names of all the authors that have written a paper with title containing the word "Monadic". | SELECT t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title LIKE "%Monadic%" | extra |
2,911 | icfp_1 | [
"authorship",
"authors",
"papers"
] | Which authors have written a paper with title containing the word "Monadic"? Return their last names. | SELECT t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title LIKE "%Monadic%" | extra |
2,912 | icfp_1 | [
"authorship",
"papers"
] | Retrieve the title of the paper that has the largest number of authors. | SELECT t2.title FROM authorship AS t1 JOIN papers AS t2 ON t1.paperid = t2.paperid WHERE t1.authorder = (SELECT max(authorder) FROM authorship) | extra |
2,913 | icfp_1 | [
"authorship",
"papers"
] | Which paper has the most authors? Give me the paper title. | SELECT t2.title FROM authorship AS t1 JOIN papers AS t2 ON t1.paperid = t2.paperid WHERE t1.authorder = (SELECT max(authorder) FROM authorship) | extra |
2,914 | icfp_1 | [
"authors"
] | What is the first name of the author with last name "Ueno"? | SELECT fname FROM authors WHERE lname = "Ueno" | easy |
2,915 | icfp_1 | [
"authors"
] | Which authors have last name "Ueno"? List their first names. | SELECT fname FROM authors WHERE lname = "Ueno" | easy |
2,916 | icfp_1 | [
"authors"
] | Find the last name of the author with first name "Amal". | SELECT lname FROM authors WHERE fname = "Amal" | easy |
2,917 | icfp_1 | [
"authors"
] | Which authors have first name "Amal"? List their last names. | SELECT lname FROM authors WHERE fname = "Amal" | easy |
2,918 | icfp_1 | [
"authors"
] | Find the first names of all the authors ordered in alphabetical order. | SELECT fname FROM authors ORDER BY fname | easy |
2,919 | icfp_1 | [
"authors"
] | Sort the first names of all the authors in alphabetical order. | SELECT fname FROM authors ORDER BY fname | easy |
2,920 | icfp_1 | [
"authors"
] | Retrieve all the last names of authors in alphabetical order. | SELECT lname FROM authors ORDER BY lname | easy |
2,921 | icfp_1 | [
"authors"
] | Give me a list of all the last names of authors sorted in alphabetical order | SELECT lname FROM authors ORDER BY lname | easy |
2,922 | icfp_1 | [
"authors"
] | Retrieve all the first and last names of authors in the alphabetical order of last names. | SELECT fname , lname FROM authors ORDER BY lname | medium |
2,923 | icfp_1 | [
"authors"
] | Sort the list of all the first and last names of authors in alphabetical order of the last names. | SELECT fname , lname FROM authors ORDER BY lname | medium |
2,924 | sakila_1 | [
"actor"
] | How many different last names do the actors and actresses have? | SELECT count(DISTINCT last_name) FROM actor | easy |
2,925 | sakila_1 | [
"actor"
] | Count the number of different last names actors have. | SELECT count(DISTINCT last_name) FROM actor | easy |
2,926 | sakila_1 | [
"actor"
] | What is the most popular first name of the actors? | SELECT first_name FROM actor GROUP BY first_name ORDER BY count(*) DESC LIMIT 1 | hard |
2,927 | sakila_1 | [
"actor"
] | Return the most common first name among all actors. | SELECT first_name FROM actor GROUP BY first_name ORDER BY count(*) DESC LIMIT 1 | hard |
2,928 | sakila_1 | [
"actor"
] | What is the most popular full name of the actors? | SELECT first_name , last_name FROM actor GROUP BY first_name , last_name ORDER BY count(*) DESC LIMIT 1 | hard |
2,929 | sakila_1 | [
"actor"
] | Return the most common full name among all actors. | SELECT first_name , last_name FROM actor GROUP BY first_name , last_name ORDER BY count(*) DESC LIMIT 1 | hard |
2,930 | sakila_1 | [
"address"
] | Which districts have at least two addresses? | SELECT district FROM address GROUP BY district HAVING count(*) >= 2 | easy |
2,931 | sakila_1 | [
"address"
] | Give the districts which have two or more addresses. | SELECT district FROM address GROUP BY district HAVING count(*) >= 2 | easy |
2,932 | sakila_1 | [
"address"
] | What is the phone number and postal code of the address 1031 Daugavpils Parkway? | SELECT phone , postal_code FROM address WHERE address = '1031 Daugavpils Parkway' | medium |
2,933 | sakila_1 | [
"address"
] | Give the phone and postal code corresponding to the address '1031 Daugavpils Parkway'. | SELECT phone , postal_code FROM address WHERE address = '1031 Daugavpils Parkway' | medium |
2,934 | sakila_1 | [
"address",
"city"
] | Which city has the most addresses? List the city name, number of addresses, and city id. | SELECT T2.city , count(*) , T1.city_id FROM address AS T1 JOIN city AS T2 ON T1.city_id = T2.city_id GROUP BY T1.city_id ORDER BY count(*) DESC LIMIT 1 | extra |
2,935 | sakila_1 | [
"address",
"city"
] | What are the city name, id, and number of addresses corresponding to the city with the most addressed? | SELECT T2.city , count(*) , T1.city_id FROM address AS T1 JOIN city AS T2 ON T1.city_id = T2.city_id GROUP BY T1.city_id ORDER BY count(*) DESC LIMIT 1 | extra |
2,936 | sakila_1 | [
"address"
] | How many addresses are in the district of California? | SELECT count(*) FROM address WHERE district = 'California' | easy |
2,937 | sakila_1 | [
"address"
] | Count the number of addressed in the California district. | SELECT count(*) FROM address WHERE district = 'California' | easy |
2,938 | sakila_1 | [
"inventory",
"film"
] | Which film is rented at a fee of 0.99 and has less than 3 in the inventory? List the film title and id. | SELECT title , film_id FROM film WHERE rental_rate = 0.99 INTERSECT SELECT T1.title , T1.film_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id GROUP BY T1.film_id HAVING count(*) < 3 | extra |
2,939 | sakila_1 | [
"inventory",
"film"
] | What are the title and id of the film which has a rental rate of 0.99 and an inventory of below 3? | SELECT title , film_id FROM film WHERE rental_rate = 0.99 INTERSECT SELECT T1.title , T1.film_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id GROUP BY T1.film_id HAVING count(*) < 3 | extra |
2,940 | sakila_1 | [
"city",
"country"
] | How many cities are in Australia? | SELECT count(*) FROM city AS T1 JOIN country AS T2 ON T1.country_id = T2.country_id WHERE T2.country = 'Australia' | medium |
2,941 | sakila_1 | [
"city",
"country"
] | Count the number of cities in Australia. | SELECT count(*) FROM city AS T1 JOIN country AS T2 ON T1.country_id = T2.country_id WHERE T2.country = 'Australia' | medium |
2,942 | sakila_1 | [
"city",
"country"
] | Which countries have at least 3 cities? | SELECT T2.country FROM city AS T1 JOIN country AS T2 ON T1.country_id = T2.country_id GROUP BY T2.country_id HAVING count(*) >= 3 | medium |
2,943 | sakila_1 | [
"city",
"country"
] | What are the countries that contain 3 or more cities? | SELECT T2.country FROM city AS T1 JOIN country AS T2 ON T1.country_id = T2.country_id GROUP BY T2.country_id HAVING count(*) >= 3 | medium |
2,944 | sakila_1 | [
"payment",
"staff"
] | Find all the payment dates for the payments with an amount larger than 10 and the payments handled by a staff person with the first name Elsa. | SELECT payment_date FROM payment WHERE amount > 10 UNION SELECT T1.payment_date FROM payment AS T1 JOIN staff AS T2 ON T1.staff_id = T2.staff_id WHERE T2.first_name = 'Elsa' | hard |
2,945 | sakila_1 | [
"payment",
"staff"
] | What are the payment dates for any payments that have an amount greater than 10 or were handled by a staff member with the first name Elsa? | SELECT payment_date FROM payment WHERE amount > 10 UNION SELECT T1.payment_date FROM payment AS T1 JOIN staff AS T2 ON T1.staff_id = T2.staff_id WHERE T2.first_name = 'Elsa' | hard |
2,946 | sakila_1 | [
"customer"
] | How many customers have an active value of 1? | SELECT count(*) FROM customer WHERE active = '1' | easy |
2,947 | sakila_1 | [
"customer"
] | Count the number of customers who are active. | SELECT count(*) FROM customer WHERE active = '1' | easy |
2,948 | sakila_1 | [
"film"
] | Which film has the highest rental rate? And what is the rate? | SELECT title , rental_rate FROM film ORDER BY rental_rate DESC LIMIT 1 | medium |
2,949 | sakila_1 | [
"film"
] | What are the title and rental rate of the film with the highest rental rate? | SELECT title , rental_rate FROM film ORDER BY rental_rate DESC LIMIT 1 | medium |
2,950 | sakila_1 | [
"film_actor",
"film"
] | Which film has the most number of actors or actresses? List the film name, film id and description. | SELECT T2.title , T2.film_id , T2.description FROM film_actor AS T1 JOIN film AS T2 ON T1.film_id = T2.film_id GROUP BY T2.film_id ORDER BY count(*) DESC LIMIT 1 | extra |
2,951 | sakila_1 | [
"film_actor",
"film"
] | What are the title, id, and description of the movie with the greatest number of actors? | SELECT T2.title , T2.film_id , T2.description FROM film_actor AS T1 JOIN film AS T2 ON T1.film_id = T2.film_id GROUP BY T2.film_id ORDER BY count(*) DESC LIMIT 1 | extra |
2,952 | sakila_1 | [
"film_actor",
"actor"
] | Which film actor (actress) starred the most films? List his or her first name, last name and actor id. | SELECT T2.first_name , T2.last_name , T2.actor_id FROM film_actor AS T1 JOIN actor AS T2 ON T1.actor_id = T2.actor_id GROUP BY T2.actor_id ORDER BY count(*) DESC LIMIT 1 | extra |
2,953 | sakila_1 | [
"film_actor",
"actor"
] | Return the full name and id of the actor or actress who starred in the greatest number of films. | SELECT T2.first_name , T2.last_name , T2.actor_id FROM film_actor AS T1 JOIN actor AS T2 ON T1.actor_id = T2.actor_id GROUP BY T2.actor_id ORDER BY count(*) DESC LIMIT 1 | extra |
2,954 | sakila_1 | [
"film_actor",
"actor"
] | Which film actors (actresses) played a role in more than 30 films? List his or her first name and last name. | SELECT T2.first_name , T2.last_name FROM film_actor AS T1 JOIN actor AS T2 ON T1.actor_id = T2.actor_id GROUP BY T2.actor_id HAVING count(*) > 30 | medium |
2,955 | sakila_1 | [
"film_actor",
"actor"
] | What are the full names of actors who had roles in more than 30 films? | SELECT T2.first_name , T2.last_name FROM film_actor AS T1 JOIN actor AS T2 ON T1.actor_id = T2.actor_id GROUP BY T2.actor_id HAVING count(*) > 30 | medium |
2,956 | sakila_1 | [
"inventory"
] | Which store owns most items? | SELECT store_id FROM inventory GROUP BY store_id ORDER BY count(*) DESC LIMIT 1 | hard |
2,957 | sakila_1 | [
"inventory"
] | What is the id of the store that has the most items in inventory? | SELECT store_id FROM inventory GROUP BY store_id ORDER BY count(*) DESC LIMIT 1 | hard |
2,958 | sakila_1 | [
"payment"
] | What is the total amount of all payments? | SELECT sum(amount) FROM payment | easy |
2,959 | sakila_1 | [
"payment"
] | Return the sum of all payment amounts. | SELECT sum(amount) FROM payment | easy |
2,960 | sakila_1 | [
"payment",
"customer"
] | Which customer, who has made at least one payment, has spent the least money? List his or her first name, last name, and the id. | SELECT T1.first_name , T1.last_name , T1.customer_id FROM customer AS T1 JOIN payment AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY sum(amount) ASC LIMIT 1 | extra |
2,961 | sakila_1 | [
"payment",
"customer"
] | What is the full name and id of the customer who has the lowest total amount of payment? | SELECT T1.first_name , T1.last_name , T1.customer_id FROM customer AS T1 JOIN payment AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY sum(amount) ASC LIMIT 1 | extra |
2,962 | sakila_1 | [
"category",
"film_category",
"film"
] | What is the genre name of the film HUNGER ROOF? | SELECT T1.name FROM category AS T1 JOIN film_category AS T2 ON T1.category_id = T2.category_id JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T3.title = 'HUNGER ROOF' | hard |
2,963 | sakila_1 | [
"category",
"film_category",
"film"
] | Return the name of the category to which the film 'HUNGER ROOF' belongs. | SELECT T1.name FROM category AS T1 JOIN film_category AS T2 ON T1.category_id = T2.category_id JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T3.title = 'HUNGER ROOF' | hard |
2,964 | sakila_1 | [
"category",
"film_category"
] | How many films are there in each category? List the genre name, genre id and the count. | SELECT T2.name , T1.category_id , count(*) FROM film_category AS T1 JOIN category AS T2 ON T1.category_id = T2.category_id GROUP BY T1.category_id | medium |
2,965 | sakila_1 | [
"category",
"film_category"
] | What are the names and ids of the different categories, and how many films are in each? | SELECT T2.name , T1.category_id , count(*) FROM film_category AS T1 JOIN category AS T2 ON T1.category_id = T2.category_id GROUP BY T1.category_id | medium |
2,966 | sakila_1 | [
"inventory",
"film"
] | Which film has the most copies in the inventory? List both title and id. | SELECT T1.title , T1.film_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id GROUP BY T1.film_id ORDER BY count(*) DESC LIMIT 1 | extra |
2,967 | sakila_1 | [
"inventory",
"film"
] | What is the title and id of the film that has the greatest number of copies in inventory? | SELECT T1.title , T1.film_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id GROUP BY T1.film_id ORDER BY count(*) DESC LIMIT 1 | extra |
2,968 | sakila_1 | [
"rental",
"inventory",
"film"
] | What is the film title and inventory id of the item in the inventory which was rented most frequently? | SELECT T1.title , T2.inventory_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id JOIN rental AS T3 ON T2.inventory_id = T3.inventory_id GROUP BY T2.inventory_id ORDER BY count(*) DESC LIMIT 1 | extra |
2,969 | sakila_1 | [
"rental",
"inventory",
"film"
] | Return the title and inventory id of the film that is rented most often. | SELECT T1.title , T2.inventory_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id JOIN rental AS T3 ON T2.inventory_id = T3.inventory_id GROUP BY T2.inventory_id ORDER BY count(*) DESC LIMIT 1 | extra |
2,970 | sakila_1 | [
"film"
] | How many languages are in these films? | SELECT count(DISTINCT language_id) FROM film | easy |
2,971 | sakila_1 | [
"film"
] | Count the number of different languages in these films. | SELECT count(DISTINCT language_id) FROM film | easy |
2,972 | sakila_1 | [
"film"
] | What are all the movies rated as R? List the titles. | SELECT title FROM film WHERE rating = 'R' | easy |
2,973 | sakila_1 | [
"film"
] | Return the titles of any movies with an R rating. | SELECT title FROM film WHERE rating = 'R' | easy |
2,974 | sakila_1 | [
"address",
"store"
] | Where is store 1 located? | SELECT T2.address FROM store AS T1 JOIN address AS T2 ON T1.address_id = T2.address_id WHERE store_id = 1 | medium |
2,975 | sakila_1 | [
"address",
"store"
] | Return the address of store 1. | SELECT T2.address FROM store AS T1 JOIN address AS T2 ON T1.address_id = T2.address_id WHERE store_id = 1 | medium |
2,976 | sakila_1 | [
"payment",
"staff"
] | Which staff handled least number of payments? List the full name and the id. | SELECT T1.first_name , T1.last_name , T1.staff_id FROM staff AS T1 JOIN payment AS T2 ON T1.staff_id = T2.staff_id GROUP BY T1.staff_id ORDER BY count(*) ASC LIMIT 1 | extra |
2,977 | sakila_1 | [
"payment",
"staff"
] | Give the full name and staff id of the staff who has handled the fewest payments. | SELECT T1.first_name , T1.last_name , T1.staff_id FROM staff AS T1 JOIN payment AS T2 ON T1.staff_id = T2.staff_id GROUP BY T1.staff_id ORDER BY count(*) ASC LIMIT 1 | extra |
2,978 | sakila_1 | [
"LANGUAGE",
"film"
] | Which language does the film AIRPORT POLLOCK use? List the language name. | SELECT T2.name FROM film AS T1 JOIN LANGUAGE AS T2 ON T1.language_id = T2.language_id WHERE T1.title = 'AIRPORT POLLOCK' | medium |
2,979 | sakila_1 | [
"LANGUAGE",
"film"
] | What is the name of the language that the film 'AIRPORT POLLOCK' is in? | SELECT T2.name FROM film AS T1 JOIN LANGUAGE AS T2 ON T1.language_id = T2.language_id WHERE T1.title = 'AIRPORT POLLOCK' | medium |
2,980 | sakila_1 | [
"store"
] | How many stores are there? | SELECT count(*) FROM store | easy |
2,981 | sakila_1 | [
"store"
] | Count the number of stores. | SELECT count(*) FROM store | easy |
2,982 | sakila_1 | [
"film"
] | How many kinds of different ratings are listed? | SELECT count(DISTINCT rating) FROM film | easy |
2,983 | sakila_1 | [
"film"
] | Count the number of different film ratings. | SELECT count(DISTINCT rating) FROM film | easy |
2,984 | sakila_1 | [
"film"
] | Which movies have 'Deleted Scenes' as a substring in the special feature? | SELECT title FROM film WHERE special_features LIKE '%Deleted Scenes%' | medium |
2,985 | sakila_1 | [
"film"
] | Return the titles of films that include 'Deleted Scenes' in their special feature section. | SELECT title FROM film WHERE special_features LIKE '%Deleted Scenes%' | medium |
2,986 | sakila_1 | [
"inventory"
] | How many items in inventory does store 1 have? | SELECT count(*) FROM inventory WHERE store_id = 1 | easy |
2,987 | sakila_1 | [
"inventory"
] | Count the number of items store 1 has in stock. | SELECT count(*) FROM inventory WHERE store_id = 1 | easy |
2,988 | sakila_1 | [
"payment"
] | When did the first payment happen? | SELECT payment_date FROM payment ORDER BY payment_date ASC LIMIT 1 | medium |
2,989 | sakila_1 | [
"payment"
] | What was the date of the earliest payment? | SELECT payment_date FROM payment ORDER BY payment_date ASC LIMIT 1 | medium |
2,990 | sakila_1 | [
"address",
"customer"
] | Where does the customer with the first name Linda live? And what is her email? | SELECT T2.address , T1.email FROM customer AS T1 JOIN address AS T2 ON T2.address_id = T1.address_id WHERE T1.first_name = 'LINDA' | medium |
2,991 | sakila_1 | [
"address",
"customer"
] | Return the address and email of the customer with the first name Linda. | SELECT T2.address , T1.email FROM customer AS T1 JOIN address AS T2 ON T2.address_id = T1.address_id WHERE T1.first_name = 'LINDA' | medium |
2,992 | sakila_1 | [
"film"
] | Find all the films longer than 100 minutes, or rated PG, except those who cost more than 200 for replacement. List the titles. | SELECT title FROM film WHERE LENGTH > 100 OR rating = 'PG' EXCEPT SELECT title FROM film WHERE replacement_cost > 200 | extra |
2,993 | sakila_1 | [
"film"
] | What are the titles of films that are either longer than 100 minutes or rated PG other than those that cost more than 200 to replace? | SELECT title FROM film WHERE LENGTH > 100 OR rating = 'PG' EXCEPT SELECT title FROM film WHERE replacement_cost > 200 | extra |
2,994 | sakila_1 | [
"rental",
"customer"
] | What is the first name and the last name of the customer who made the earliest rental? | SELECT T1.first_name , T1.last_name FROM customer AS T1 JOIN rental AS T2 ON T1.customer_id = T2.customer_id ORDER BY T2.rental_date ASC LIMIT 1 | hard |
2,995 | sakila_1 | [
"rental",
"customer"
] | Return the full name of the customer who made the first rental. | SELECT T1.first_name , T1.last_name FROM customer AS T1 JOIN rental AS T2 ON T1.customer_id = T2.customer_id ORDER BY T2.rental_date ASC LIMIT 1 | hard |
2,996 | sakila_1 | [
"rental",
"customer",
"staff"
] | What is the full name of the staff member who has rented a film to a customer with the first name April and the last name Burns? | SELECT DISTINCT T1.first_name , T1.last_name FROM staff AS T1 JOIN rental AS T2 ON T1.staff_id = T2.staff_id JOIN customer AS T3 ON T2.customer_id = T3.customer_id WHERE T3.first_name = 'APRIL' AND T3.last_name = 'BURNS' | hard |
2,997 | sakila_1 | [
"rental",
"customer",
"staff"
] | Return the full name of the staff who provided a customer with the first name April and the last name Burns with a film rental. | SELECT DISTINCT T1.first_name , T1.last_name FROM staff AS T1 JOIN rental AS T2 ON T1.staff_id = T2.staff_id JOIN customer AS T3 ON T2.customer_id = T3.customer_id WHERE T3.first_name = 'APRIL' AND T3.last_name = 'BURNS' | hard |
2,998 | sakila_1 | [
"customer"
] | Which store has most the customers? | SELECT store_id FROM customer GROUP BY store_id ORDER BY count(*) DESC LIMIT 1 | hard |
2,999 | sakila_1 | [
"customer"
] | Return the id of the store with the most customers. | SELECT store_id FROM customer GROUP BY store_id ORDER BY count(*) DESC LIMIT 1 | hard |
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