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Write SQL query to solve given problem: How many actors have starred in the film ACADEMY DINOSAUR?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T1.actor_id) FROM film_actor AS T1 INNER JOIN film AS T2 ON T1.film_id = T2.film_id WHERE T2.title = 'ACADEMY DINOSAUR' |
Write SQL query to solve given problem: Please list the full names of all the actors that have starred in the film ACADEMY DINOSAUR.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.first_name, T1.last_name FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T3.title = 'ACADEMY DINOSAUR' |
Write SQL query to solve given problem: Among the films starring PENELOPE GUINESS, how many of them are released in 2006?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T2.film_id) FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T3.release_year = 2006 AND T1.first_name = 'PENELOPE' AND T1.last_name = 'GUINESS' |
Write SQL query to solve given problem: Please give the title of the film starring PENELOPE GUINESS and has the highest replacement cost.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T3.title FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T1.first_name = 'PENELOPE' AND T1.last_name = 'GUINESS' ORDER BY T3.replacement_cost DESC LIMIT 1 |
Write SQL query to solve given problem: Please list the full names of all the actors that have starred in the film with the highest replacement cost.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT first_name, last_name FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id ORDER BY T3.replacement_cost DESC LIMIT 1 |
Write SQL query to solve given problem: Among the films starring PENELOPE GUINESS, how many of them are in English?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T3.film_id) FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id INNER JOIN language AS T4 ON T3.language_id = T4.language_id WHERE T4.name = 'English' AND T1.first_name = 'PENELOPE' AND T1.last_name = 'GUINESS' |
Write SQL query to solve given problem: What is the title of the film with the longest duration time and stars PENELOPE GUINESS?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T3.title FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T1.first_name = 'PENELOPE' AND T1.last_name = 'GUINESS' ORDER BY T3.length DESC LIMIT 1 |
Write SQL query to solve given problem: Please list the titles of all the films in the category of "Horror".. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.title FROM film AS T1 INNER JOIN film_category AS T2 ON T1.film_id = T2.film_id INNER JOIN category AS T3 ON T2.category_id = T3.category_id WHERE T3.name = 'Horror' |
Write SQL query to solve given problem: How many films are there under the category of "Horror"?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T1.film_id) FROM film_category AS T1 INNER JOIN category AS T2 ON T1.category_id = T2.category_id WHERE T2.name = 'Horror' |
Write SQL query to solve given problem: Please list the titles of all the films under the category of "Horror" and has a rental rate of $2.99.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.title FROM film AS T1 INNER JOIN film_category AS T2 ON T1.film_id = T2.film_id INNER JOIN category AS T3 ON T2.category_id = T3.category_id WHERE T3.name = 'Horror' AND T1.rental_rate = 2.99 |
Write SQL query to solve given problem: For how many times has the customer RUTH MARTINEZ rented a film?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T2.rental_id) FROM customer AS T1 INNER JOIN rental AS T2 ON T1.customer_id = T2.customer_id WHERE T1.first_name = 'RUTH' AND T1.last_name = 'MARTINEZ' |
Write SQL query to solve given problem: Please list the titles of all the films that the customer RUTH MARTINEZ has rented.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T4.title FROM customer AS T1 INNER JOIN rental AS T2 ON T1.customer_id = T2.customer_id INNER JOIN inventory AS T3 ON T2.inventory_id = T3.inventory_id INNER JOIN film AS T4 ON T3.film_id = T4.film_id WHERE T1.first_name = 'RUTH' AND T1.last_name = 'MARTINEZ' |
Write SQL query to solve given problem: Among the films that the customer RUTH MARTINEZ has rented, how many of them are released in 2006?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T1.customer_id) FROM customer AS T1 INNER JOIN rental AS T2 ON T1.customer_id = T2.customer_id INNER JOIN inventory AS T3 ON T2.inventory_id = T3.inventory_id INNER JOIN film AS T4 ON T3.film_id = T4.film_id WHERE T4.release_year = 2006 AND T1.first_name = 'RUTH' AND T1.last_name = 'MARTINEZ' |
Write SQL query to solve given problem: Among the films that the customer RUTH MARTINEZ has rented, what is the title of the one with the highest replacement cost?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T4.title FROM customer AS T1 INNER JOIN rental AS T2 ON T1.customer_id = T2.customer_id INNER JOIN inventory AS T3 ON T2.inventory_id = T3.inventory_id INNER JOIN film AS T4 ON T3.film_id = T4.film_id WHERE T1.first_name = 'RUTH' AND T1.last_name = 'MARTINEZ' ORDER BY T4.replacement_cost DESC LIMIT 1 |
Write SQL query to solve given problem: Please list the full names of all the customers who have rented the film with the highest replacement cost.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.first_name, T1.last_name FROM customer AS T1 INNER JOIN rental AS T2 ON T1.customer_id = T2.customer_id INNER JOIN inventory AS T3 ON T2.inventory_id = T3.inventory_id INNER JOIN film AS T4 ON T3.film_id = T4.film_id ORDER BY T4.replacement_cost DESC LIMIT 1 |
Write SQL query to solve given problem: How many films rented to the customer RUTH MARTINEZ were returned in August, 2005?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T1.customer_id) FROM customer AS T1 INNER JOIN rental AS T2 ON T1.customer_id = T2.customer_id WHERE T1.first_name = 'RUTH' AND T1.last_name = 'MARTINEZ' AND STRFTIME('%m',T2.return_date) = '8' AND STRFTIME('%Y', T2.return_date) = '2005' |
Write SQL query to solve given problem: Please give the full name of the customer that have rented the most films.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T.first_name, T.last_name FROM ( SELECT T1.first_name, T1.last_name, COUNT(T2.rental_id) AS num FROM customer AS T1 INNER JOIN rental AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.first_name, T1.last_name ) AS T ORDER BY T.num DESC LIMIT 1 |
Write SQL query to solve given problem: Among the customers who have rented the film ACADEMY DINOSAUR, how many of them are active?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T1.customer_id) FROM customer AS T1 INNER JOIN rental AS T2 ON T1.customer_id = T2.customer_id INNER JOIN inventory AS T3 ON T2.inventory_id = T3.inventory_id INNER JOIN film AS T4 ON T3.film_id = T4.film_id WHERE T1.active = 1 AND T4.title = 'ACADEMY DINOSAUR' |
Write SQL query to solve given problem: Which film is rented for the most times by the customers? Please give its title.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T.title FROM ( SELECT T1.title, COUNT(T3.rental_id) AS num FROM film AS T1 INNER JOIN inventory AS T2 ON T1.film_id = T2.film_id INNER JOIN rental AS T3 ON T2.inventory_id = T3.inventory_id GROUP BY T1.title ) AS T ORDER BY T.num DESC LIMIT 1 |
Write SQL query to solve given problem: Which customer has rented more movies, RUTH MARTINEZ or LINDA WILLIAMS?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T.first_name, T.last_name FROM ( SELECT T1.first_name, T1.last_name, COUNT(T1.customer_id) AS num FROM customer AS T1 INNER JOIN rental AS T2 ON T1.customer_id = T2.customer_id WHERE (T1.first_name = 'RUTH' AND T1.last_name = 'MARTINEZ') OR (T1.first_name = 'LINDA' AND T1.last_name = 'WILLIAMS') GROUP BY T1.first_name, T1.last_name ) AS T ORDER BY T.num DESC LIMIT 1 |
Write SQL query to solve given problem: Among all the films starring PENELOPE GUINESS, what is the title of the one with the highest rental price per day?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T3.title FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T1.first_name = 'PENELOPE' AND T1.last_name = 'GUINESS' ORDER BY T3.rental_rate / T3.rental_duration DESC LIMIT 1 |
Write SQL query to solve given problem: What is the average replacement cost of the films under the category of "Horror"?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT AVG(T3.replacement_cost) FROM film_category AS T1 INNER JOIN category AS T2 ON T1.category_id = T2.category_id INNER JOIN film AS T3 ON T1.film_id = T3.film_id WHERE T2.name = 'Horror' |
Write SQL query to solve given problem: Among all films that the customer RUTH MARTINEZ has rented, what is the percentage of it being a Music film?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT CAST(SUM(IIF(T3.name = 'Music', 1, 0)) AS REAL) * 100 / COUNT(T1.film_id) FROM film AS T1 INNER JOIN film_category AS T2 ON T1.film_id = T2.film_id INNER JOIN category AS T3 ON T2.category_id = T3.category_id INNER JOIN inventory AS T4 ON T1.film_id = T4.film_id INNER JOIN customer AS T5 ON T4.store_id = T5.store_id INNER JOIN rental AS T6 ON T4.inventory_id = T6.inventory_id WHERE T5.first_name = 'RUTH' AND T5.last_name = 'MARTINEZ' |
Write SQL query to solve given problem: What is the average duration time of the films starring PENELOPE GUINESS?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT AVG(T3.length) FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T1.first_name = 'PENELOPE' AND T1.last_name = 'GUINESS' |
Write SQL query to solve given problem: What is Diane Collins' email address?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT email FROM customer WHERE first_name = 'DIANE' AND last_name = 'COLLINS' |
Write SQL query to solve given problem: Give the number of inactive customers.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(customer_id) FROM customer WHERE active = 0 |
Write SQL query to solve given problem: Who is the owner of email address "[email protected]"? Give the full name.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT first_name, last_name FROM customer WHERE email = '[email protected]' |
Write SQL query to solve given problem: Give the postal code for the address No.65.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT postal_code FROM address WHERE address_id = 65 |
Write SQL query to solve given problem: State the number of addresses in the Nordrhein-Westfalen district.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(address_id) FROM address WHERE district = 'Nordrhein-Westfalen' |
Write SQL query to solve given problem: What is the phone number of address No.72?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT phone FROM address WHERE address_id = '72' |
Write SQL query to solve given problem: State the number of films that are 178 minutes long.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(film_id) FROM film WHERE length = '178' |
Write SQL query to solve given problem: Tell the special features of the film Uprising Uptown.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT special_features FROM film WHERE title = 'UPRISING UPTOWN' |
Write SQL query to solve given problem: What is the description of the film Artist Coldblooded?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT description FROM film WHERE title = 'ARTIST COLDBLOODED' |
Write SQL query to solve given problem: Give the detailed address for store No.2.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.address, T1.address2, T1.district FROM address AS T1 INNER JOIN store AS T2 ON T1.address_id = T2.address_id WHERE T2.store_id = 2 |
Write SQL query to solve given problem: Which continent is the mother country of Clarksville city in?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.country FROM country AS T1 INNER JOIN city AS T2 ON T1.country_id = T2.country_id WHERE T2.city = 'Clarksville' |
Write SQL query to solve given problem: How many actors played a role in the 2006 film whose rental duration is 7 days, rental rate is 4.99 and is 98 minutes duration?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T1.actor_id) FROM film_actor AS T1 INNER JOIN film AS T2 ON T1.film_id = T2.film_id WHERE T2.release_year = 2006 AND T2.rental_duration = 7 AND T2.rental_duration = 4.99 AND T2.length = 98 |
Write SQL query to solve given problem: The actor Dan Harris played in a 77 minute film with replacement cost of 9.99, what was the rating for that film?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T3.rating FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T1.first_name = 'DAN' AND T1.last_name = 'HARRIS' AND T3.length = 77 AND T3.replacement_cost = '9.99' |
Write SQL query to solve given problem: How many films did actor Daryl Wahlberg appear in?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T1.film_id) FROM film_actor AS T1 INNER JOIN actor AS T2 ON T1.actor_id = T2.actor_id WHERE T2.first_name = 'DARYL' AND T2.last_name = 'WAHLBERG' |
Write SQL query to solve given problem: Sherri Rhodes rented a film at 12:27:27 on 2005/7/28, when did she/he return that film?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T2.return_date FROM customer AS T1 INNER JOIN rental AS T2 ON T1.customer_id = T2.customer_id WHERE T1.first_name = 'SHERRI' AND T1.last_name = 'RHODES' AND T2.rental_date = '2005-07-28 12:27:27' |
Write SQL query to solve given problem: Give the name of the manager staff for store No.1.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.first_name, T1.last_name FROM staff AS T1 INNER JOIN store AS T2 ON T1.store_id = T2.store_id WHERE T2.store_id = 1 |
Write SQL query to solve given problem: State the address location of store No.1.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.address, T1.address2, T1.district FROM address AS T1 INNER JOIN store AS T2 ON T1.address_id = T2.address_id WHERE T2.store_id = 1 |
Write SQL query to solve given problem: Where does the staff Jon Stephens live?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.address, T1.address2 FROM address AS T1 INNER JOIN staff AS T2 ON T1.address_id = T2.address_id WHERE T2.first_name = 'Jon' AND T2.last_name = 'Stephens' |
Write SQL query to solve given problem: How many addresses are there in Woodridge city?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T1.address_id) FROM address AS T1 INNER JOIN city AS T2 ON T1.city_id = T2.city_id WHERE T2.city = 'Woodridge' |
Write SQL query to solve given problem: How many films are in English?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T1.film_id) FROM film AS T1 INNER JOIN language AS T2 ON T1.language_id = T2.language_id WHERE T2.name = 'English' |
Write SQL query to solve given problem: Give the address location of Heather Morris.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.address FROM address AS T1 INNER JOIN customer AS T2 ON T1.address_id = T2.address_id WHERE T2.first_name = 'HEATHER' AND T2.last_name = 'MORRIS' |
Write SQL query to solve given problem: Give the email address of the person who lives in "1411 Lillydale Drive".. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T2.email FROM address AS T1 INNER JOIN staff AS T2 ON T1.address_id = T2.address_id WHERE T1.address = '1411 Lillydale Drive' |
Write SQL query to solve given problem: How much money did the customer No.297 pay for the rental which happened at 12:27:27 on 2005/7/28?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.amount FROM payment AS T1 INNER JOIN rental AS T2 ON T1.rental_id = T2.rental_id WHERE T2.rental_date = '2005-07-28 12:27:27' AND T2.customer_id = 297 |
Write SQL query to solve given problem: Which category does the film Working Microcosmos belong to?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T3.name FROM film AS T1 INNER JOIN film_category AS T2 ON T1.film_id = T2.film_id INNER JOIN category AS T3 ON T2.category_id = T3.category_id WHERE T1.title = 'WORKING MICROCOSMOS' |
Write SQL query to solve given problem: Give the number of documentary films.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T1.film_id) FROM film_category AS T1 INNER JOIN category AS T2 ON T1.category_id = T2.category_id WHERE T2.name = 'Documentary' |
Write SQL query to solve given problem: State the name of the category which has the most number of films.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T.name FROM ( SELECT T2.name, COUNT(T1.film_id) AS num FROM film_category AS T1 INNER JOIN category AS T2 ON T1.category_id = T2.category_id GROUP BY T2.name ) AS T ORDER BY T.num DESC LIMIT 1 |
Write SQL query to solve given problem: Give the name of the film for inventory No.3479.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.title FROM film AS T1 INNER JOIN inventory AS T2 ON T1.film_id = T2.film_id WHERE T2.inventory_id = 3479 |
Write SQL query to solve given problem: What is the percentage more for the rental payment for store No.2 than store No.1?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT CAST((SUM(IIF(T2.store_id = 2, T1.amount, 0)) - SUM(IIF(T2.store_id = 1, T1.amount, 0))) AS REAL) * 100 / SUM(IIF(T2.store_id = 1, T1.amount, 0)) FROM payment AS T1 INNER JOIN customer AS T2 ON T1.customer_id = T2.customer_id INNER JOIN store AS T3 ON T2.store_id = T3.store_id |
Write SQL query to solve given problem: How many times is the number of Indian cities than Italian cities?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT CAST(SUM(IIF(T1.country = 'India', 1, 0)) AS REAL) / SUM(IIF(T1.country = 'Italy', 1, 0)) FROM country AS T1 INNER JOIN city AS T2 ON T1.country_id = T2.country_id |
Write SQL query to solve given problem: How many times is the number of films Gina DeGeneres acted in than Penelope Guinness?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT CAST(SUM(IIF(T2.first_name = 'GINA' AND T2.last_name = 'DEGENERES', 1, 0)) AS REAL) * 100 / SUM(IIF(T2.first_name = 'PENELOPE' AND T2.last_name = 'GUINESS', 1, 0)) FROM film_actor AS T1 INNER JOIN actor AS T2 ON T1.actor_id = T2.actor_id |
Write SQL query to solve given problem: In 2006, how many restricted films were released?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(film_id) FROM film WHERE rating = 'R' AND release_year = 2006 |
Write SQL query to solve given problem: How many actors starred in the film id 508?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(actor_id) FROM film_actor WHERE film_id = 508 |
Write SQL query to solve given problem: What are the special features for the film "Smoochy Control"?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT special_features FROM film WHERE title = 'SMOOCHY CONTROL' |
Write SQL query to solve given problem: How many customers paid over the amount of 10 on August 2005?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(customer_id) FROM payment WHERE SUBSTR(payment_date, 1, 7) LIKE '2005-08' |
Write SQL query to solve given problem: List the names of the films that are more than 180 minutes long.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT title FROM film WHERE length > 180 |
Write SQL query to solve given problem: How much is the total rental payment for the first 10 rentals?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT SUM(amount) FROM payment WHERE rental_id BETWEEN 1 AND 10 |
Write SQL query to solve given problem: What are the full names of all the active employees?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT first_name, last_name FROM staff WHERE active = 1 |
Write SQL query to solve given problem: Who is the staff manager in store id 2?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT manager_staff_id FROM store WHERE store_id = 2 |
Write SQL query to solve given problem: How many rentals were returned on 5/27/2005?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(rental_id) FROM rental WHERE rental_date = '2005-05-27' |
Write SQL query to solve given problem: What are the names of the movies which Laura Brody starred in?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T3.title FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T1.first_name = 'Laura' AND T1.last_name = 'Brody' |
Write SQL query to solve given problem: List the name of the films that can only be found in store id 2.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.title FROM film AS T1 INNER JOIN inventory AS T2 ON T1.film_id = T2.film_id WHERE T2.store_id = 2 |
Write SQL query to solve given problem: What is the full name of the customer who rented movies for 7 consecutive days?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.first_name, T1.last_name FROM customer AS T1 INNER JOIN ( SELECT customer_id, COUNT(*) AS num_days FROM ( SELECT *, date(days, '-' || rn || ' day') AS results FROM ( SELECT customer_id, days, row_number() OVER (PARTITION BY customer_id ORDER BY days) AS rn FROM ( SELECT DISTINCT customer_id, date(rental_date) AS days FROM rental ) ) ) GROUP BY customer_id, results HAVING num_days = 7 ) AS T2 ON T1.customer_id = T2.customer_id |
Write SQL query to solve given problem: How many films are categorized as horror?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T1.film_id) FROM film_category AS T1 INNER JOIN category AS T2 ON T1.category_id = T2.category_id WHERE T2.name = 'Horror' |
Write SQL query to solve given problem: What is the name of the most rented movie?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T.title FROM ( SELECT T1.title, COUNT(T3.rental_id) AS num FROM film AS T1 INNER JOIN inventory AS T2 ON T1.film_id = T2.film_id INNER JOIN rental AS T3 ON T2.inventory_id = T3.inventory_id GROUP BY T1.title ) AS T ORDER BY T.num DESC LIMIT 1 |
Write SQL query to solve given problem: What is the most common special features of science-fiction movies?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.special_features FROM film AS T1 INNER JOIN film_category AS T2 ON T1.film_id = T2.film_id INNER JOIN category AS T3 ON T2.category_id = T3.category_id WHERE T3.name = 'sci-fi' ORDER BY T1.special_features DESC LIMIT 1 |
Write SQL query to solve given problem: What is the full name of the actor who starred in most movies?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T.first_name, T.last_name FROM ( SELECT T2.first_name, T2.last_name, COUNT(T1.film_id) AS num FROM film_actor AS T1 INNER JOIN actor AS T2 ON T1.actor_id = T2.actor_id GROUP BY T2.first_name, T2.last_name ) AS T ORDER BY T.num DESC LIMIT 1 |
Write SQL query to solve given problem: Among the films with a rental duration of 7 days, how many are comedies?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T1.film_id) FROM film AS T1 INNER JOIN film_category AS T2 ON T1.film_id = T2.film_id INNER JOIN category AS T3 ON T2.category_id = T3.category_id WHERE T1.rental_duration = 7 AND T3.name = 'Comedy' |
Write SQL query to solve given problem: Who is the staff manager of the store with the most non-active customers?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T.first_name, T.last_name FROM ( SELECT T3.first_name, T3.last_name, COUNT(T1.customer_id) AS num FROM customer AS T1 INNER JOIN store AS T2 ON T1.store_id = T2.store_id INNER JOIN staff AS T3 ON T2.store_id = T3.store_id WHERE T1.active = 0 GROUP BY T3.first_name, T3.last_name ) AS T ORDER BY T.num DESC LIMIT 1 |
Write SQL query to solve given problem: What is the rental price per day of the most expensive children's film?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.rental_rate FROM film AS T1 INNER JOIN film_category AS T2 ON T1.film_id = T2.film_id INNER JOIN category AS T3 ON T2.category_id = T3.category_id WHERE T3.name = 'Children' ORDER BY T1.rental_rate / T1.rental_duration DESC LIMIT 1 |
Write SQL query to solve given problem: What is the complete address of store id 1?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T3.address, T3.address2, T3.district FROM country AS T1 INNER JOIN city AS T2 ON T1.country_id = T2.country_id INNER JOIN address AS T3 ON T2.city_id = T3.city_id INNER JOIN store AS T4 ON T3.address_id = T4.address_id WHERE T4.store_id = 1 |
Write SQL query to solve given problem: How many customers are from the city of Lethbridge?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T3.customer_id) FROM city AS T1 INNER JOIN address AS T2 ON T1.city_id = T2.city_id INNER JOIN customer AS T3 ON T2.address_id = T3.address_id WHERE T1.city = 'Lethbridge' |
Write SQL query to solve given problem: How many cities are there in the United States?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T2.city) FROM country AS T1 INNER JOIN city AS T2 ON T1.country_id = T2.country_id WHERE T1.country = 'United States' |
Write SQL query to solve given problem: List the names of the customers from India.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T4.first_name, T4.last_name FROM country AS T1 INNER JOIN city AS T2 ON T1.country_id = T2.country_id INNER JOIN address AS T3 ON T2.city_id = T3.city_id INNER JOIN customer AS T4 ON T3.address_id = T4.address_id WHERE T1.country = 'India' |
Write SQL query to solve given problem: Among the classic movies, how many movies have a rental rate of less than 1?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T1.film_id) FROM film_category AS T1 INNER JOIN category AS T2 ON T1.category_id = T2.category_id INNER JOIN film AS T3 ON T1.film_id = T3.film_id WHERE T3.rental_rate < 1 AND T2.name = 'Classics' |
Write SQL query to solve given problem: What is the full name of the customer who rented the highest number of movies of all time?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T.first_name, T.last_name FROM ( SELECT T2.first_name, T2.last_name, COUNT(T1.rental_id) AS num FROM rental AS T1 INNER JOIN customer AS T2 ON T1.customer_id = T2.customer_id GROUP BY T2.first_name, T2.last_name ) AS T ORDER BY T.num DESC LIMIT 1 |
Write SQL query to solve given problem: How many times was "Blanket Beverly" rented?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T3.rental_id) FROM film AS T1 INNER JOIN inventory AS T2 ON T1.film_id = T2.film_id INNER JOIN rental AS T3 ON T2.inventory_id = T3.inventory_id WHERE T1.title = 'Blanket Beverly' |
Write SQL query to solve given problem: What is the full name of the actor who has the highest number of restricted films?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T.first_name, T.last_name FROM ( SELECT T1.first_name, T1.last_name, COUNT(T2.film_id) AS num FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T3.rating = 'R' GROUP BY T1.first_name, T1.last_name ) AS T ORDER BY T.num DESC LIMIT 1 |
Write SQL query to solve given problem: Who are the top 5 actors with the highest number of films? List their full names and calculate the average number of films for each of the actors.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T.first_name, T.last_name, num FROM ( SELECT T1.first_name, T1.last_name, COUNT(T2.film_id) AS num FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id GROUP BY T1.first_name, T1.last_name ) AS T ORDER BY T.num DESC LIMIT 5 |
Write SQL query to solve given problem: List the actors' IDs who have "KILMER" as last name.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT actor_id FROM actor WHERE last_name = 'KILMER' |
Write SQL query to solve given problem: List down the films titles with the lowest replacement cost under the general audiences rating.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT title FROM film WHERE replacement_cost = ( SELECT MIN(replacement_cost) FROM film ) |
Write SQL query to solve given problem: Among the films with the longest duration, list any five title with their descriptions and special features.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT title, description, special_features FROM film WHERE length = ( SELECT MAX(length) FROM film ) LIMIT 5 |
Write SQL query to solve given problem: How many films rented on 26th May, 2005 were returned on 30th May, 2005?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(DISTINCT rental_id) FROM rental WHERE date(rental_date) BETWEEN '2005-05-26' AND '2005-05-30' |
Write SQL query to solve given problem: Calculate the average payment amount per customer.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT AVG(amount) FROM payment GROUP BY customer_id |
Write SQL query to solve given problem: What is the name and email of the staff in store ID 2?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT first_name, last_name, email FROM staff WHERE store_id = 2 |
Write SQL query to solve given problem: How many percent of customers were inactive?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT CAST(SUM(IIF(active = 0, 1, 0)) AS REAL) * 100 / COUNT(customer_id) FROM customer |
Write SQL query to solve given problem: What is the description and film title of ID 996?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT description, title FROM film_text WHERE film_id = 996 |
Write SQL query to solve given problem: Calculate customers' total payment amount in August, 2005.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT SUM(amount) FROM payment WHERE SUBSTR(payment_date, 1, 7) = '2005-08' |
Write SQL query to solve given problem: List down the film titles performed by Emily Dee.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T3.title FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T1.first_name = 'Emily' AND T1.last_name = 'Dee' |
Write SQL query to solve given problem: List down the actors' full names who performed in "CHOCOLATE DUCK" film.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T3.first_name, T3.last_name FROM film_actor AS T1 INNER JOIN film AS T2 ON T1.film_id = T2.film_id INNER JOIN actor AS T3 ON T1.actor_id = T3.actor_id WHERE T2.title = 'CHOCOLATE DUCK' |
Write SQL query to solve given problem: How many films in the horror category were included in PG-13-rated?. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT COUNT(T1.film_id) FROM film AS T1 INNER JOIN film_category AS T2 ON T1.film_id = T2.film_id INNER JOIN category AS T3 ON T2.category_id = T3.category_id WHERE T3.name = 'Horror' AND T1.rating = 'PG-13' |
Write SQL query to solve given problem: Distinguish the films performed by Judy Dean according to category.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T5.name FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id INNER JOIN film_category AS T4 ON T2.film_id = T4.film_id INNER JOIN category AS T5 ON T4.category_id = T5.category_id WHERE T1.first_name = 'Judy' AND T1.last_name = 'Dean' |
Write SQL query to solve given problem: Write down any five film names under the documentary category.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.title FROM film AS T1 INNER JOIN film_category AS T2 ON T1.film_id = T2.film_id INNER JOIN category AS T3 ON T2.category_id = T3.category_id WHERE T3.name = 'Documentary' LIMIT 5 |
Write SQL query to solve given problem: Mention the language of Untouchables Sunrise film and calculate its rental cost per day.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T2.name, T1.replacement_cost / T1.rental_duration AS cost FROM film AS T1 INNER JOIN language AS T2 ON T1.language_id = T2.language_id WHERE T1.title = 'UNTOUCHABLES SUNRISE' |
Write SQL query to solve given problem: List the films' titles which were rented on 24th May,2005.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T1.title FROM film AS T1 INNER JOIN inventory AS T2 ON T1.film_id = T2.film_id INNER JOIN rental AS T3 ON T2.inventory_id = T3.inventory_id WHERE SUBSTR(T3.rental_date, 1, 10) = '2005-05-24' |
Write SQL query to solve given problem: List the films' titles which were rented by Brian Wyman in July, 2005.. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T4.title FROM customer AS T1 INNER JOIN rental AS T2 ON T1.customer_id = T2.customer_id INNER JOIN inventory AS T3 ON T2.inventory_id = T3.inventory_id INNER JOIN film AS T4 ON T3.film_id = T4.film_id WHERE T1.first_name = 'BRIAN' AND T1.last_name = 'WYMAN' AND STRFTIME('%Y', T2.rental_date) = '2005' AND STRFTIME('%m',T2.rental_date) = '7' |
Write SQL query to solve given problem: Write down the inventories' IDs and actors' names of "STREETCAR INTENTIONS".. Keep the solution inside sql tag ```sql [SQL-Query] ``` | movie_3 | SELECT T4.inventory_id, T1.first_name, T1.last_name FROM actor AS T1 INNER JOIN film_actor AS T2 ON T1.actor_id = T2.actor_id INNER JOIN film AS T3 ON T2.film_id = T3.film_id INNER JOIN inventory AS T4 ON T2.film_id = T4.film_id WHERE T3.title = 'STREETCAR INTENTIONS' |
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