lab-sql-subqueries

solutions.sql

@@ -1 +1,127 @@
--- Add you solution queries below:
+# Conducts the tasks requested in the SQL subqueries lab assignment.
+USE sakila;
+
+# 1 - Determines the number of copies of Hunchback Impossible that exist in the inventory system.
+SELECT f.title,
+       COUNT(i.inventory_id) AS number_of_copies
+FROM film AS f
+JOIN inventory AS i 
+    ON f.film_id = i.film_id
+WHERE f.title = 'Hunchback Impossible';
+
+# 2 - Lists all films whose length is longer than the average length of all films
+SELECT title,
+       length
+FROM film
+WHERE length > (SELECT AVG(length) FROM film);
+
+# 3 - Displays all actors who appear in the film Alone Trip
+SELECT a.first_name,
+       a.last_name
+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 - Identifies all movies categorized as family films
+SELECT f.title
+FROM film AS f
+JOIN film_category AS fc
+    ON f.film_id = fc.film_id
+JOIN category AS c
+    ON fc.category_id = c.category_id
+WHERE c.name = 'Family';
+
+#5a - Obtains the name and email from Canada customers (subquery approach)
+SELECT c.first_name,
+       c.last_name,
+       c.email
+FROM customer AS c
+WHERE c.address_id IN (
+    SELECT a.address_id
+    FROM address AS a
+    WHERE a.city_id IN (
+        SELECT ci.city_id
+        FROM city AS ci
+        WHERE ci.country_id = (
+            SELECT co.country_id
+            FROM country AS co
+            WHERE co.country = 'Canada'
+        )
+    )
+);
+
+# 5b - Obtains the name and email from Canada customers (join approach)
+SELECT c.first_name,
+       c.last_name,
+       c.email
+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';
+
+# 6a Finds the films starred by the most prolific actor (shows the actor_id)
+SELECT fa.actor_id,
+       COUNT(*) AS film_count
+FROM film_actor AS fa
+GROUP BY fa.actor_id
+ORDER BY film_count DESC
+LIMIT 1;
+
+# 6b - finds all the films that the actor in 6a starred in
+# Uses the actor_id obtained in 6a
+SELECT f.title
+FROM film AS f
+JOIN film_actor AS fa
+    ON f.film_id = fa.film_id
+WHERE fa.actor_id = 107; -- actor_id from step 6a
+
+# 7a - Finds the most profitable customer
+# Most profitable is defined as the customer with the largest total payments
+SELECT p.customer_id,
+       SUM(p.amount) AS total_spent
+FROM payment AS p
+GROUP BY p.customer_id
+ORDER BY total_spent DESC
+LIMIT 1;
+
+# 7b - finds the films rented by the most profitable customer
+# Uses the customer_id from 7a
+SELECT DISTINCT f.title
+FROM rental AS r
+JOIN inventory AS i
+    ON r.inventory_id = i.inventory_id
+JOIN film AS f
+    ON i.film_id = f.film_id
+WHERE r.customer_id = 526;  -- customer_id from step 7a
+
+/* 8 - Finds the customer_id(client_id) and the total amount spent for those who spent
+more than the average of what each client spends */
+SELECT t.customer_id,
+       t.total_spent
+FROM
+(
+    -- Subquery: total spent by each customer
+    SELECT p.customer_id,
+           SUM(p.amount) AS total_spent
+    FROM payment AS p
+    GROUP BY p.customer_id
+) AS t
+WHERE t.total_spent > (
+    -- Compare to average of total_spent across all customers
+    SELECT AVG(t2.total_spent)
+    FROM (
+        SELECT p2.customer_id,
+               SUM(p2.amount) AS total_spent
+        FROM payment AS p2
+        GROUP BY p2.customer_id
+    ) AS t2
+);