Dan Answers

.github/workflows/snorkell-auto-documentation.yml

@@ -0,0 +1,19 @@
+# This workflow will improvise current file with AI genereated documentation and Create new PR
+
+name: Penify - Revolutionizing Documentation on GitHub
+
+on:
+  push:
+    branches: ["master"]
+  workflow_dispatch:
+
+jobs:
+  Documentation:
+    runs-on: ubuntu-latest
+    steps:
+    - name: Penify DocGen Client
+      uses: SingularityX-ai/snorkell-documentation-client@v1.0.0
+      with:
+        client_id: ${{ secrets.SNORKELL_CLIENT_ID }}
+        api_key: ${{ secrets.SNORKELL_API_KEY }}
+        branch_name: "master"

solutions.sql

@@ -1 +1,180 @@
--- Add you solution queries below:
+
+
+-- How many copies of the film Hunchback Impossible exist in the inventory system?
+SELECT 
+    f.title AS `Film Title`,
+    COUNT(i.inventory_id) AS `Total Copies`
+FROM 
+    film f
+JOIN 
+    inventory i ON f.film_id = i.film_id
+WHERE 
+    f.title = 'Hunchback Impossible'
+GROUP BY 
+    f.title;
+
+
+-- List all films whose length is longer than the average of all the films.
+
+SELECT 
+    title AS `Film Title`,
+    length AS `Film Length`
+FROM 
+    film
+WHERE 
+    length > (SELECT AVG(length) FROM film)
+ORDER BY 
+    length DESC;
+
+
+-- Use subqueries to display all actors who appear in the film Alone Trip.
+
+SELECT 
+    a.actor_id AS `Actor ID`,
+    a.first_name AS `First Name`,
+    a.last_name AS `Last Name`
+FROM 
+    actor a
+WHERE 
+    a.actor_id IN (
+        SELECT fa.actor_id
+        FROM film_actor fa
+        JOIN film f ON fa.film_id = f.film_id
+        WHERE f.title = 'Alone Trip'
+    );
+
+
+-- Identify all movies categorized as family films.
+SELECT 
+    f.title AS `Film Title`,
+    c.name AS `Category`
+FROM 
+    film f
+JOIN 
+    film_category fc ON f.film_id = fc.film_id
+JOIN 
+    category c ON fc.category_id = c.category_id
+WHERE 
+    c.name = 'Family';
+
+
+
+-- Get name and email from customers from Canada using subqueries. Do the same with joins. 
+-- Note that to create a join, you will have to identify the correct tables with their primary keys and foreign keys, 
+-- that will help you get the relevant information.
+SELECT 
+    first_name AS `First Name`,
+    last_name AS `Last Name`,
+    email AS `Email`
+FROM 
+    customer
+WHERE 
+    address_id IN (
+        SELECT address_id
+        FROM address
+        WHERE city_id IN (
+            SELECT city_id
+            FROM city
+            WHERE country_id = (
+                SELECT country_id
+                FROM country
+                WHERE country = 'Canada'
+            )
+        )
+    );
+
+-- with joins
+SELECT 
+    c.first_name AS `First Name`,
+    c.last_name AS `Last Name`,
+    c.email AS `Email`
+FROM 
+    customer c
+JOIN 
+    address a ON c.address_id = a.address_id
+JOIN 
+    city ci ON a.city_id = ci.city_id
+JOIN 
+    country co ON ci.country_id = co.country_id
+WHERE 
+    co.country = 'Canada';
+
+
+
+
+
+
+-- Which are films starred by the most prolific actor? 
+-- Most prolific actor is defined as the actor that has acted in the most number of films. 
+-- First you will have to find the most prolific actor and then use that actor_id to find the different films that he/she starred.
+SELECT 
+    fa.actor_id AS `Actor ID`,
+    a.first_name AS `First Name`,
+    a.last_name AS `Last Name`,
+    COUNT(fa.film_id) AS `Total Films`
+FROM 
+    film_actor fa
+JOIN 
+    actor a ON fa.actor_id = a.actor_id
+GROUP BY 
+    fa.actor_id, a.first_name, a.last_name
+ORDER BY 
+    `Total Films` DESC
+LIMIT 1;
+
+
+-- Films rented by most profitable customer. 
+-- You can use the customer table and payment table to find the most profitable customer 
+-- ie the customer that has made the largest sum of payments
+
+SELECT 
+    f.title AS `Film Title`
+FROM 
+    rental r
+JOIN 
+    inventory i ON r.inventory_id = i.inventory_id
+JOIN 
+    film f ON i.film_id = f.film_id
+WHERE 
+    r.customer_id = (
+        SELECT 
+            c.customer_id
+        FROM 
+            customer c
+        JOIN 
+            payment p ON c.customer_id = p.customer_id
+        GROUP BY 
+            c.customer_id
+        ORDER BY 
+            SUM(p.amount) DESC
+        LIMIT 1
+    );
+
+
+
+
+-- Get 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.
+
+SELECT 
+    customer_id AS `Customer ID`,
+    SUM(amount) AS `Total Amount Spent`
+FROM 
+    payment
+GROUP BY 
+    customer_id
+HAVING 
+    SUM(amount) > (
+        SELECT AVG(total_amount)
+        FROM (
+            SELECT 
+                customer_id, 
+                SUM(amount) AS total_amount
+            FROM 
+                payment
+            GROUP BY 
+                customer_id
+        ) AS customer_totals
+    );
+
+