Solved Lab

lab agrega transform.sql

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+
+/*
+You need to use SQL built-in functions to gain insights relating to the duration of movies:
+
+1.1 Determine the shortest and longest movie durations and name the values as max_duration and min_duration.
+
+1.2. Express the average movie duration in hours and minutes. Don't use decimals.
+
+Hint: Look for floor and round functions.
+
+*/
+
+USE sakila;
+-- To test table
+SELECT * FROM film as f;
+
+-- 1.1
+SELECT MIN(f.length) AS 'Minimum_amount', MAX(f.length) AS 'Maximum_amount'
+FROM film as f;
+-- 1.2
+SELECT 
+    FLOOR(AVG(f.length) / 60) AS horas,
+    ROUND(AVG(f.length) % 60) AS minutos
+FROM film AS f;
+-- 2.1
+SELECT 
+    MIN(rental_date) AS primera_fecha,
+    MAX(rental_date) AS ultima_fecha,
+    DATEDIFF(MAX(rental_date), MIN(rental_date)) AS dias_operando
+FROM rental;
+-- 2.2
+SELECT 
+    rental_id,
+    rental_date,
+    customer_id,
+    inventory_id,
+    staff_id,
+    return_date,
+    MONTHNAME(rental_date) AS rental_month,
+    DAYNAME(rental_date) AS rental_weekday
+FROM rental
+ORDER BY rental_date
+LIMIT 20;
+-- 2.3
+SELECT 
+    rental_id,
+    rental_date,
+    customer_id,
+    inventory_id,
+    staff_id,
+    DAYNAME(rental_date) AS rental_weekday,
+    CASE 
+        WHEN WEEKDAY(rental_date) IN (5, 6) THEN 'weekend'
+        ELSE 'workday'
+    END AS DAY_TYPE
+FROM rental
+ORDER BY rental_date
+LIMIT 20;
+-- 1.1
+SELECT 
+    COUNT(*) AS total_films
+FROM film;
+-- 1.2
+SELECT 
+    rating,
+    COUNT(*) AS number_of_films
+FROM film
+GROUP BY rating;
+-- 1.3
+SELECT 
+    rating,
+    COUNT(*) AS number_of_films
+FROM film
+GROUP BY rating
+ORDER BY number_of_films DESC;
+-- 2.1
+SELECT 
+    rating,
+    ROUND(AVG(length), 2) AS mean_duration
+FROM film
+GROUP BY rating
+ORDER BY mean_duration DESC;
+-- 2.2
+SELECT 
+    rating,
+    ROUND(AVG(length), 2) AS mean_duration
+FROM film
+GROUP BY rating
+HAVING AVG(length) > 120
+ORDER BY mean_duration DESC;
+
+
+
+
+