From c288928895e57940081373546dd86f6ade30818d Mon Sep 17 00:00:00 2001 From: Ghazal Hassanzadeh Date: Sun, 7 Dec 2025 16:21:01 +0100 Subject: [PATCH 1/2] Solved lab --- sql_basic_queries_lab_results.sql | 52 +++++++++++++++++++++++++++++++ 1 file changed, 52 insertions(+) create mode 100644 sql_basic_queries_lab_results.sql diff --git a/sql_basic_queries_lab_results.sql b/sql_basic_queries_lab_results.sql new file mode 100644 index 0000000..0518a10 --- /dev/null +++ b/sql_basic_queries_lab_results.sql @@ -0,0 +1,52 @@ +-- challenge 1: Display all available tables in the Sakila database. +USE sakila; +SHOW TABLES; + +-- challenge 2: Retrieve all the data from the tables actor, film and customer. +SELECT * FROM sakila.actor; +SELECT * FROM sakila.film; +SELECT * FROM sakila.customer; + +-- I combined all the 3 tables together and I got a lack of memory error. so I had to run them seperately. + +-- challenge 3.1: Titles of all films from the film table +SELECT title FROM film; + +-- challenge 3.2: List of languages used in films, with the column aliased as language from the language table. +SELECT name FROM language AS language; + +-- challenge 3.3: List of first names of all employees from the staff table +SELECT first_name FROM staff; + +-- challenge 4: Retrieve unique release years. +SELECT DISTINCT release_year FROM film; + +-- challenge 5.1: Determine the number of stores that the company has. +SELECT COUNT(*) AS number_of_stores FROM store; + +-- challenge 5.2: Determine the number of employees that the company has. +SELECT COUNT(*) AS number_of_employees FROM staff; + +-- challenge 5.3: Determine how many films are available for rent and how many have been rented. +SELECT (SELECT COUNT(*) FROM inventory) AS available_for_rent, +(SELECT COUNT(*) FROM rental) AS rented; + +-- challenge 5.4: Determine the number of distinct last names of the actors in the database. +SELECT Distinct last_name AS distinct_actors_last_names FROM actor; + +-- challenge 6: Retrieve the 10 longest films. +SELECT title, length FROM film +order by length DESC +LIMIT 10; + +-- challenge 7.1: Retrieve all actors with the first name "SCARLETT". +SELECT * FROM actor +WHERE first_name = "SCARLETT"; + +-- challenge 7.2: Retrieve all movies that have ARMAGEDDON in their title and have a duration longer than 100 minutes. +SELECT title, length FROM film +where title like "%ARMAGEDDON%" AND length > 100; + +-- challenge 7.3: Determine the number of films that include Behind the Scenes content +SELECT count(*) AS films_with_behind_the_scenes FROM film +where special_features like "%Behind the Scenes%"; From c22f71bea87a320e9630af2c1bbc6d855ecaf64a Mon Sep 17 00:00:00 2001 From: Ghazal Hassanzadeh Date: Sun, 7 Dec 2025 16:29:31 +0100 Subject: [PATCH 2/2] Solved Lab --- sql_basic_queries_lab_results.sql | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/sql_basic_queries_lab_results.sql b/sql_basic_queries_lab_results.sql index 0518a10..9ac9923 100644 --- a/sql_basic_queries_lab_results.sql +++ b/sql_basic_queries_lab_results.sql @@ -22,17 +22,17 @@ SELECT first_name FROM staff; SELECT DISTINCT release_year FROM film; -- challenge 5.1: Determine the number of stores that the company has. -SELECT COUNT(*) AS number_of_stores FROM store; +SELECT COUNT(DISTINCT store_id) AS number_of_stores FROM store; -- challenge 5.2: Determine the number of employees that the company has. -SELECT COUNT(*) AS number_of_employees FROM staff; +SELECT COUNT(DISTINCT staff_id) AS number_of_employees FROM staff; -- challenge 5.3: Determine how many films are available for rent and how many have been rented. SELECT (SELECT COUNT(*) FROM inventory) AS available_for_rent, (SELECT COUNT(*) FROM rental) AS rented; -- challenge 5.4: Determine the number of distinct last names of the actors in the database. -SELECT Distinct last_name AS distinct_actors_last_names FROM actor; +SELECT COUNT(DISTINCT last_name) AS distinct_last_names FROM actor; -- challenge 6: Retrieve the 10 longest films. SELECT title, length FROM film