Solved lab

solved_lab.sql

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+use sakila;
+
+-- Determine the shortest and longest movie durations and name the values as max_duration and min_duration
+
+select max(length) as "max_duration" , min(length) as "min_duration"
+from film;
+
+select floor(avg(length))
+from film;
+
+-- You need to gain insights related to rental dates:
+
+select datediff(max(rental_date), min(rental_date))
+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,
+       DAYNAME(rental_date) AS rental_weekday,
+       CASE 
+           WHEN DAYOFWEEK(rental_date) IN (1, 7) THEN 'weekend'
+           ELSE 'workday'
+       END AS DAY_TYPE
+FROM rental
+LIMIT 20;
+
+-- You need to ensure that customers can easily access information about the movie collection. To achieve this, 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.
+-- Please note that even if there are currently no null values in the rental duration column, the query should still be written to handle such cases in the future.
+-- Hint: Look for the IFNULL() function.
+
+select ifnull(rental_duration, "Not Available") as "rental_duration", title
+from film;
+
+-- bonus
+
+select concat(first_name, " ", last_name) as "full_name", left(email, 3) as "short_email"
+from customer
+order by last_name asc;
+
+-- Next, you need to analyze the films in the collection to gain some more insights. Using the film table, determine: 
+
+select count(film_id)
+from film;
+
+select count(rating), rating
+from film
+group by rating;
+
+select count(rating), rating
+from film
+group by rating
+order by count(rating) desc;
+
+-- Using the film table, determine: .
+
+select round(avg(length), 2), rating
+from film
+group by rating
+order by round(avg(length), 2) desc;
+
+select round(avg(length), 2), rating
+from film
+group by rating
+having avg(length) > 120;
+
+-- bonus
+
+select count(last_name), last_name
+from actor
+group by last_name
+having count(last_name) = 1;