Solved lab

lab-sql-subqueries.sql

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+-- ==================================================
+-- SETTING UP THE DATABASE
+-- ==================================================
+
+USE sakila;
+
+-- ==================================================
+-- CHALLENGE
+-- ==================================================
+
+-- 1. Number of copies of "Hunchback Impossible" that exist in the inventory system.
+SELECT COUNT(*) AS copies
+FROM inventory i
+INNER JOIN film f ON i.film_id = f.film_id
+WHERE f.title = 'Hunchback Impossible';
+-- 6 copies.
+
+-- 2. Films whose length is longer than the average length of all films.
+-- The inner query computes the average once; the outer query reuses that single number as
+-- the threshold -- same pattern as the loan-amount subquery from the class script.
+SELECT title, length
+FROM film
+WHERE length > (SELECT AVG(length) FROM film)
+ORDER BY length DESC;
+-- 489 films qualify (out of 1000 total).
+
+-- 3. Actors who appear in "Alone Trip", using a subquery.
+SELECT first_name, last_name
+FROM actor
+WHERE actor_id IN (
+    SELECT fa.actor_id
+    FROM film_actor fa
+    INNER JOIN film f ON fa.film_id = f.film_id
+    WHERE f.title = 'Alone Trip'
+)
+ORDER BY last_name;
+
+-- ==================================================
+-- BONUS
+-- ==================================================
+
+-- 4. All films categorized as "Family".
+SELECT title
+FROM film
+WHERE film_id IN (
+    SELECT fc.film_id
+    FROM film_category fc
+    INNER JOIN category c ON fc.category_id = c.category_id
+    WHERE c.name = 'Family'
+)
+ORDER BY title;
+-- 69 family films.
+
+-- 5. Name and email of customers from Canada -- once with a subquery, once with joins.
+-- Subquery version:
+SELECT first_name, last_name, email
+FROM customer
+WHERE address_id IN (
+    SELECT a.address_id
+    FROM address a
+    INNER JOIN city ci ON a.city_id = ci.city_id
+    INNER JOIN country co ON ci.country_id = co.country_id
+    WHERE co.country = 'Canada'
+);
+
+-- Join version (address -> city -> country is the only path to "country" from a customer,
+-- there's no shortcut column on customer/address itself):
+SELECT c.first_name, c.last_name, c.email
+FROM customer c
+INNER JOIN address a ON c.address_id = a.address_id
+INNER JOIN city ci ON a.city_id = ci.city_id
+INNER JOIN country co ON ci.country_id = co.country_id
+WHERE co.country = 'Canada';
+-- Both return the same 5 customers.
+
+-- 6. Films starred by the most prolific actor (most films acted in).
+-- Step 1 as its own subquery: which actor_id has the highest film count.
+SELECT actor_id, COUNT(*) AS film_count
+FROM film_actor
+GROUP BY actor_id
+ORDER BY film_count DESC
+LIMIT 1;
+-- actor_id 107, 42 films.
+
+-- Step 2: use that actor_id (as a subquery, not hardcoded) to pull their films.
+SELECT f.title
+FROM film f
+INNER JOIN film_actor fa ON f.film_id = fa.film_id
+WHERE fa.actor_id = (
+    SELECT actor_id
+    FROM film_actor
+    GROUP BY actor_id
+    ORDER BY COUNT(*) DESC
+    LIMIT 1
+)
+ORDER BY f.title;
+-- 42 films back, matching the film_count from step 1 -- confirms the subquery picked the
+-- same actor consistently.
+
+-- 7. Films rented by the most profitable customer (highest total payments).
+SELECT f.title
+FROM film f
+INNER JOIN inventory i ON f.film_id = i.film_id
+INNER JOIN rental r ON i.inventory_id = r.inventory_id
+WHERE r.customer_id = (
+    SELECT customer_id
+    FROM payment
+    GROUP BY customer_id
+    ORDER BY SUM(amount) DESC
+    LIMIT 1
+)
+ORDER BY f.title;
+-- customer_id 526 ($221.55 total) -- same customer (Karl Seal) that came out on top in the
+-- Temporary Tables/Views/CTEs lab's customer summary report, a useful cross-check that both
+-- labs' logic agrees.
+
+-- 8. client_id and total_amount_spent for clients who spent more than the average total
+-- spent per client. Built as a CTE, since "total spent per client" is needed twice --
+-- once as the actual output, once inside the AVG() that filters it -- same reuse-a-named-
+-- result idea as the class CTE examples.
+WITH customer_totals AS (
+    SELECT customer_id, SUM(amount) AS total_amount_spent
+    FROM payment
+    GROUP BY customer_id
+)
+SELECT customer_id, ROUND(total_amount_spent, 2) AS total_amount_spent
+FROM customer_totals
+WHERE total_amount_spent > (SELECT AVG(total_amount_spent) FROM customer_totals)
+ORDER BY total_amount_spent DESC;
+-- 285 out of 599 customers spend above the average -- expected with typically right-skewed
+-- spending data (a handful of high spenders pull the average above where "most" people sit).