And solutions for SQL data aggregation and transformation lab

lab-agg-tranf.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,
+    FLOOR(AVG(length) % 60) AS avg_minutes
+FROM film;
+
+SELECT 
+    DATEDIFF(MAX(rental_date), MIN(rental_date)) AS operating_days
+FROM rental;
+
+SELECT 
+    rental_id,
+    rental_date,
+    MONTH(rental_date) AS rental_month,
+    DAYNAME(rental_date) AS rental_weekday
+FROM rental
+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
+LIMIT 20;    
+
+SELECT
+    title,
+    IFNULL(rental_duration, '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;
+
+
+SELECT COUNT(*) AS total_films
+FROM film;
+
+SELECT  rating, COUNT(*) AS film_count
+FROM film
+GROUP BY rating;
+
+SELECT rating, COUNT(*) AS film_count
+FROM film
+GROUP BY rating
+ORDER BY film_count DESC;
+
+SELECT 
+    rating, 
+    ROUND(AVG(length), 2) AS avg_duration
+FROM film
+GROUP BY rating
+ORDER BY avg_duration DESC;
+
+SELECT 
+    rating, 
+    ROUND(AVG(length), 2) AS avg_duration
+FROM film
+GROUP BY rating
+HAVING AVG(length) > 120;
+
+SELECT last_name
+FROM actor
+GROUP BY last_name
+HAVING COUNT(*) = 1;
+
+
+
+
+
+
+