SolvedLab

sakila-subqueries.sql

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+-- Use this 'sakila' Database
+USE sakila;
+SELECT * FROM inventory;
+-- 1. Determine the number of copies of the film "Hunchback Impossible" that exist in the inventory system.
+SELECT 
+	f.title,
+    f.film_id,
+    COUNT(i.film_id) AS number_of_copies
+FROM film AS f
+JOIN inventory AS i ON i.film_id=f.film_id
+WHERE f.title = "Hunchback Impossible"
+GROUP BY f.title, f.film_id;
+-- 2. List all films whose length is longer than the average length of all the films in the Sakila database.
+SELECT 
+    title,
+    length
+FROM film
+GROUP BY title, length
+HAVING length > (SELECT ROUND(AVG(length),2) FROM film)
+ORDER BY length;
+-- 3. Use a subquery to display all actors who appear in the film "Alone Trip".
+SELECT
+    CONCAT(a.first_name,' ',a.last_name) AS actor
+FROM
+	actor AS a
+WHERE a.actor_id IN(
+	SELECT fa.actor_id
+    FROM film_actor AS fa
+    JOIN film AS f ON fa.film_id=f.film_id
+	WHERE f.title= "Alone Trip"
+);
+-- 4. Sales have been lagging among young families, and you want to target family movies for a promotion. Identify all movies categorized as family films.
+SELECT DISTINCT
+    f.title AS titles_for_family
+FROM film AS f
+WHERE f.film_id IN (
+    SELECT fc.film_id
+    FROM film_category AS fc
+    JOIN category AS c ON fc.category_id = c.category_id
+    WHERE c.name IN ('Family', 'Children')
+);
+-- 5.Retrieve the name and email of customers from Canada using both subqueries and joins. To use joins, you will need to identify the relevant tables and their primary and foreign keys.
+-- Using joints
+SELECT 
+	CONCAT(c.first_name,' ',c.last_name) AS customer_name,
+    c.email,
+    co.country
+FROM customer AS c
+JOIN address AS a ON c.address_id=a.address_id
+JOIN city AS ci ON a.city_id=ci.city_id
+JOIN country AS co ON ci.country_id=co.country_id
+WHERE co.country='Canada';
+-- Using subqueries
+SELECT 
+	CONCAT(c.first_name,' ',c.last_name) AS customer_name,
+    c.email
+FROM customer AS c
+WHERE address_id IN (
+	SELECT a.address_id
+    FROM address AS a
+    JOIN city AS ci ON a.city_id=ci.city_id
+    JOIN country AS co ON ci.country_id=co.country_id
+    WHERE co.country='Canada'
+);
+-- 6. Determine which films were starred by the most prolific actor in the Sakila database. A prolific actor is defined as the actor who has acted in the most number of films. First, you will need to find the most prolific actor and then use that actor_id to find the different films that he or she starred in.
+SELECT
+    CONCAT(a.first_name, ' ', a.last_name) AS actor_name,
+    f.title
+FROM film AS f
+JOIN film_actor AS fa ON f.film_id = fa.film_id
+JOIN actor AS a ON fa.actor_id = a.actor_id
+WHERE fa.actor_id IN (
+    SELECT
+        actor_id
+    FROM (
+        SELECT
+            actor_id,
+            COUNT(film_id) AS film_count
+        FROM
+            film_actor
+        GROUP BY
+            actor_id
+    ) AS counts
+    WHERE film_count = (
+        SELECT MAX(film_count) 
+        FROM ( 
+            SELECT COUNT(film_id) AS film_count 
+            FROM film_actor 
+            GROUP BY actor_id 
+        ) AS subquery 
+    )
+)
+ORDER BY actor_name, f.title;
+-- Find the films rented by the most profitable customer in the Sakila database. You can use the customer and payment tables to find the most profitable customer, i.e., the customer who has made the largest sum of payments.
+SELECT
+    f.title AS titles,
+    SUM(p.amount) AS total_amount
+FROM film AS f
+JOIN inventory AS i ON f.film_id = i.film_id
+JOIN rental AS r ON i.inventory_id = r.inventory_id
+JOIN payment AS p ON p.rental_id = r.rental_id
+WHERE r.customer_id = (
+    SELECT customer_id
+    FROM payment
+    GROUP BY customer_id
+    HAVING SUM(amount) = (
+        SELECT MAX(total_payment)
+        FROM ( 
+            SELECT SUM(amount) AS total_payment
+            FROM payment
+            GROUP BY customer_id
+        ) AS subquery
+    )
+)
+GROUP BY f.title
+ORDER BY total_amount DESC;
+-- 8. Retrieve the client_id and the total_amount_spent of those clients who spent more than the average of the total_amount spent by each client. You can use subqueries to accomplish this.
+SELECT 
+    customer_id,
+    SUM(amount) AS total_amount_spent
+FROM payment
+GROUP BY customer_id
+HAVING SUM(amount) > (
+    SELECT AVG(total_spent)
+    FROM (
+        SELECT SUM(amount) AS total_spent
+        FROM payment
+        GROUP BY customer_id
+    ) AS subquery
+);