Aggregation_sql-solved

Aggregation_transformation-solved.sql

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+## CHALLENGE 1
+## 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
+SELECT max(length) AS longest_duration, min(length) AS shortest_duration FROM film;
+#1.2 Express the average movie duration in hours and minutes. Don't use decimals
+SELECT floor(AVG(length)/60) AS Average_hour_duration, round(AVG(length))%60 AS average_minutes FROM film;
+
+# You need to gain insights related to rental dates:
+#2.1 Calculate the number of days that the company has been operating
+SELECT datediff(max(rental_date),min(rental_date)) 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 *, DATE_FORMAT(rental_date,"%m") AS month, DATE_FORMAT(rental_date,"%d") AS day FROM rental;
+#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 *,
+CASE 
+		WHEN DAYOFWEEK(rental_date) IN (1, 7) THEN 'Weekend'
+		ELSE 'Weekday'
+END AS DayType
+FROM rental;
+#2.4 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
+SELECT *, IFNULL(rental_duration, "Not available") FROM film
+ORDER BY rental_duration ASC;
+
+#4 Bonus: Marketing team
+SELECT concat(first_name," ",last_name) AS full_name, LEFT(email,3) AS trunc_email FROM customer
+ORDER BY last_name ASC;
+
+## CHALLENGE 2
+## Next, you need to analyze the films in the collection to gain some more insights. Using the film table, determine:
+
+#1.1 The total number of films that have been released
+SELECT count(*) FROM film;
+#1.2 The number of films for each rating
+SELECT rating, count(*) AS number_of_films FROM film
+GROUP BY rating;
+#1.3 The number of films for each rating 
+SELECT rating, count(*) AS number_of_films FROM film
+GROUP BY rating
+order by number_of_films DESC;
+
+## Using the film table, determine:
+#2.1 The mean film duration for each rating
+SELECT rating, count(*) AS number_of_films, round(AVG(length),2) AS mean_film_duration FROM film
+GROUP BY rating
+order by number_of_films DESC;
+#2.2 The mean duration of over 2hours
+SELECT rating, count(*) AS number_of_films, floor(AVG(length)/60) AS mean_film_duration_hr FROM film
+GROUP BY rating
+having mean_film_duration_hr>=2
+order by number_of_films DESC;
+
+## Bonus
+## Unrepeated last_names
+SELECT last_name FROM actor
+group by last_name
+having count(last_name)=1;