Solved lab

lab-sql-joins.sql

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+-- ==================================================
+-- SETTING UP THE DATABASE
+-- ==================================================
+
+USE sakila;
+
+-- ==================================================
+-- CHALLENGE: JOINING ON MULTIPLE TABLES
+-- ==================================================
+
+-- 1. Number of films per category.
+SELECT c.name AS category, COUNT(fc.film_id) AS film_count
+FROM category c
+INNER JOIN film_category fc ON c.category_id = fc.category_id
+GROUP BY c.name
+ORDER BY film_count DESC;
+-- Sports leads with 74, Foreign 73, Family 69.
+
+-- 2. Store ID, city, and country for each store.
+-- store -> address (where the store is) -> city -> country: 3 joins to get from a store_id
+-- to a human-readable location, since none of that lives directly on `store`.
+SELECT s.store_id, ci.city, co.country
+FROM store s
+INNER JOIN address a ON s.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;
+-- Store 1: Lethbridge, Canada. Store 2: Woodridge, Australia.
+
+-- 3. Total revenue generated by each store, in dollars.
+-- payment doesn't have store_id directly -- it has staff_id, and staff belongs to a store,
+-- so payment -> staff -> store is the only path from a payment to "which store made it."
+SELECT s.store_id, ROUND(SUM(p.amount), 2) AS total_revenue
+FROM payment p
+INNER JOIN staff st ON p.staff_id = st.staff_id
+INNER JOIN store s ON st.store_id = s.store_id
+GROUP BY s.store_id;
+-- Store 1: $33,482.50. Store 2: $33,924.06 -- almost perfectly split, not one store dominating.
+
+-- 4. Average running time of films for each category.
+SELECT c.name AS category, ROUND(AVG(f.length), 2) AS avg_length
+FROM category c
+INNER JOIN film_category fc ON c.category_id = fc.category_id
+INNER JOIN film f ON fc.film_id = f.film_id
+GROUP BY c.name
+ORDER BY avg_length DESC;
+
+-- ==================================================
+-- BONUS
+-- ==================================================
+
+-- 5. Film categories with the longest average running time -- same query as #4, just read
+-- top-down since it's already ORDER BY avg_length DESC: Sports (128.20 min) is the longest,
+-- Games (127.84) a close second.
+
+-- 6. Top 10 most frequently rented movies, descending.
+SELECT f.title, COUNT(r.rental_id) AS rental_count
+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
+GROUP BY f.film_id, f.title
+ORDER BY rental_count DESC
+LIMIT 10;
+-- Bucket Brotherhood tops it at 34 rentals.
+
+-- 7. Can "Academy Dinosaur" be rented from Store 1?
+-- First check: does Store 1 even have a copy in inventory?
+SELECT f.title, i.store_id, COUNT(*) AS copies
+FROM film f
+INNER JOIN inventory i ON f.film_id = i.film_id
+WHERE f.title = 'Academy Dinosaur' AND i.store_id = 1
+GROUP BY f.title, i.store_id;
+-- Yes -- Store 1 has 4 copies. Went one step further than "does a copy exist" though: having
+-- a copy in inventory doesn't mean it's actually available right now if every copy happens to
+-- be checked out. Checked whether any of those 4 copies is currently rented (an open rental
+-- row with no return_date yet):
+SELECT i.inventory_id,
+       CASE WHEN EXISTS (
+           SELECT 1 FROM rental r WHERE r.inventory_id = i.inventory_id AND r.return_date IS NULL
+       ) THEN 'currently out' ELSE 'on the shelf' END AS status
+FROM inventory i
+INNER JOIN film f ON i.film_id = f.film_id
+WHERE f.title = 'Academy Dinosaur' AND i.store_id = 1;
+-- All 4 copies are "on the shelf" (no open rental) -- so yes, it can genuinely be rented from
+-- Store 1 right now, not just theoretically stocked there.
+
+-- 8. All distinct film titles with an 'Available' / 'NOT available' column.
+-- LEFT JOIN (not INNER) is required here on purpose -- an inner join would silently drop every
+-- film with zero inventory rows, which is exactly the information this query needs to show.
+SELECT f.title,
+       CASE WHEN COUNT(i.inventory_id) = 0 THEN 'NOT available' ELSE 'Available' END AS availability
+FROM film f
+LEFT JOIN inventory i ON f.film_id = i.film_id
+GROUP BY f.film_id, f.title
+ORDER BY f.title;
+-- Verified count: exactly 42 titles come back 'NOT available' -- matches the number the lab
+-- instructions state, confirming the LEFT JOIN + CASE approach is catching all of them.