Solved lab

Challenge 1.sql

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+USE sakila;
+SELECT 
+  MIN(length) AS min_duration,
+  MAX(length) AS max_duration
+FROM film;
+SELECT
+  FLOOR(AVG(length) / 60) AS avg_hours,
+  (ROUND(AVG(length)) - 60 * FLOOR(AVG(length) / 60)) AS avg_minutes
+FROM film;
+SELECT 
+  DATEDIFF(MAX(rental_date), MIN(rental_date)) AS days_operating
+FROM rental;
+SELECT 
+  rental_id,
+  rental_date,
+  DATE_FORMAT(rental_date, '%M') AS rental_month,
+  DATE_FORMAT(rental_date, '%W') AS rental_weekday
+FROM rental
+ORDER BY rental_date
+LIMIT 20;
+SELECT 
+  rental_id,
+  rental_date,
+  CASE 
+    WHEN DAYOFWEEK(rental_date) IN (1,7) THEN 'weekend'
+    ELSE 'workday'
+  END AS day_type
+FROM rental
+ORDER BY rental_date
+LIMIT 20;
+SELECT 
+  title,
+  IFNULL(CAST(rental_duration AS CHAR), 'Not Available') AS rental_duration
+FROM film
+ORDER BY title ASC;
+SELECT
+  CONCAT(first_name, ' ', last_name) AS full_name,
+  SUBSTRING(email, 1, 3)           AS email_prefix
+FROM customer
+ORDER BY last_name ASC, first_name ASC;

Challenge 2.sql

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+SELECT COUNT(*) AS total_films
+FROM film;
+SELECT rating, COUNT(*) AS num_films
+FROM film
+GROUP BY rating;
+SELECT rating, COUNT(*) AS num_films
+FROM film
+GROUP BY rating
+ORDER BY num_films DESC;
+SELECT 
+  rating,
+  ROUND(AVG(length), 2) AS mean_duration
+FROM film
+GROUP BY rating
+ORDER BY mean_duration DESC;
+SELECT 
+  rating,
+  ROUND(AVG(length), 2) AS mean_duration
+FROM film
+GROUP BY rating
+HAVING AVG(length) > 120
+ORDER BY mean_duration DESC;
+SELECT last_name
+FROM actor
+GROUP BY last_name
+HAVING COUNT(*) = 1
+ORDER BY last_name;