Solved Lab

sql-transformation-aggregation.sql

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+USE sakila;
+
+-- CHALLENGE 1
+-- 1.1 Determine the shortest and longest movie durations and name the values as max_duration and min_duration
+SELECT * FROM film;
+SELECT MAX(length) AS max_duration
+FROM film;
+SELECT MIN(length) AS min_duration
+FROM film;
+
+-- 1.2. Express the average movie duration in hours and minutes. Don't use decimals
+SELECT 
+  FLOOR(AVG(length) / 60) AS hours,
+  FLOOR(MOD(AVG(length), 60)) AS minutes
+FROM film;
+
+-- 2.1 Calculate the number of days that the company has been operating
+SELECT 
+  DATEDIFF(MAX(rental_date), MIN(rental_date)) AS days_operating
+FROM rental;
+
+-- 2.2 Retrieve rental information and add two additional columns to show the month and weekday of the rental. Return 20 rows of results
+SELECT * FROM rental;
+SELECT *,
+  DATE_FORMAT(rental_date, '%M') AS month,
+  DATE_FORMAT(rental_date, '%W') AS day
+FROM rental
+LIMIT 20;
+
+-- 2.3 Bonus: Retrieve rental information and add an additional column called DAY_TYPE with values 'weekend' or 'workday', depending on the day of the week.
+SELECT * FROM rental;
+SELECT rental_date,
+CASE 
+     WHEN DATE_FORMAT(rental_date, '%W') = 'Saturday' OR 'Sunday' THEN 'weekend'
+    ELSE 'weekday'
+END AS 'DAY_TYPE'
+FROM rental;
+
+-- 3 Retrieve the film titles and their rental duration. If any rental duration value is NULL, replace it with the string 'Not Available'. Sort the results of the film title in ascending order.
+SELECT title,
+  CASE 
+    WHEN rental_duration IS NULL THEN 'Not Available'
+    ELSE rental_duration
+  END AS rental_duration
+FROM film
+ORDER BY title ASC;
+
+-- 4 Retrieve the concatenated first and last names of customers, along with the first 3 characters of their email address;
+-- Sort by last name in ascending order
+SELECT 
+  CONCAT(first_name, ' ', last_name) AS full_name,
+  LEFT(email, 3) AS email_3
+FROM customer
+ORDER BY last_name ASC;
+
+
+-- CHALLENGE 2
+
+-- 1.1 Find the total number of films that have been released
+SELECT * FROM film;
+SELECT COUNT(*) AS total_released FROM film
+WHERE release_year IS NOT NULL;
+
+-- 1.2 The number of films for each rating
+SELECT * FROM film;
+SELECT rating, COUNT(*) AS total_films
+FROM film
+GROUP BY rating;
+
+-- 1.3 The number of films for each rating, sorting the results in descending order of the number of films
+SELECT rating, COUNT(*) AS total_films
+FROM film
+GROUP BY rating
+ORDER BY total_films DESC;
+
+-- 2.1 The mean film duration for each rating, and sort the results in descending order of the mean duration. Round off the average lengths to two decimal places
+SELECT rating, ROUND(AVG(length),2) AS mean_duration
+FROM film
+GROUP BY rating
+ORDER BY mean_duration DESC;
+
+-- 2.2 Identify which ratings have a mean duration of over two hours
+SELECT rating, ROUND(AVG(length), 2) AS mean_duration
+FROM film
+GROUP BY rating
+HAVING mean_duration > 120;
+
+-- 3 Determine which last names are not repeated in the table actor
+SELECT last_name
+FROM actor
+GROUP BY last_name
+HAVING COUNT(*) = 1;