SQL exercises on movie Database: Find the first and last name of all the directors with number of genres movies they directed with genres name, and arranged the result alphabetically with the first and last name of the director
SQL movie Database: Join Exercise-6 with Solution
6. From the following table, write a SQL query to find the directors who have directed films in a variety of genres. Group the result set on director first name, last name and generic title. Sort the result-set in ascending order by director first name and last name. Return director first name, last name and number of genres movies.
Sample table: director
dir_id | dir_fname | dir_lname
--------+----------------------+----------------------
201 | Alfred | Hitchcock
202 | Jack | Clayton
203 | David | Lean
204 | Michael | Cimino
205 | Milos | Forman
206 | Ridley | Scott
207 | Stanley | Kubrick
208 | Bryan | Singer
209 | Roman | Polanski
210 | Paul | Thomas Anderson
211 | Woody | Allen
212 | Hayao | Miyazaki
213 | Frank | Darabont
214 | Sam | Mendes
215 | James | Cameron
216 | Gus | Van Sant
217 | John | Boorman
218 | Danny | Boyle
219 | Christopher | Nolan
220 | Richard | Kelly
221 | Kevin | Spacey
222 | Andrei | Tarkovsky
223 | Peter | Jackson
Sample table: movie_direction
dir_id | mov_id
--------+--------
201 | 901
202 | 902
203 | 903
204 | 904
205 | 905
206 | 906
207 | 907
208 | 908
209 | 909
210 | 910
211 | 911
212 | 912
213 | 913
214 | 914
215 | 915
216 | 916
217 | 917
218 | 918
219 | 919
220 | 920
218 | 921
215 | 922
221 | 923
Sample table: genres
gen_id | gen_title --------+---------------------- 1001 | Action 1002 | Adventure 1003 | Animation 1004 | Biography 1005 | Comedy 1006 | Crime 1007 | Drama 1008 | Horror 1009 | Music 1010 | Mystery 1011 | Romance 1012 | Thriller 1013 | War
Sample table: movie_genres
mov_id | gen_id
--------+--------
922 | 1001
917 | 1002
903 | 1002
912 | 1003
911 | 1005
908 | 1006
913 | 1006
926 | 1007
928 | 1007
918 | 1007
921 | 1007
902 | 1008
923 | 1009
907 | 1010
927 | 1010
901 | 1010
914 | 1011
906 | 1012
904 | 1013
Sample Solution:
-- Selecting director first names, last names, genre titles, and the count of genre titles
-- Using the 'director' table
SELECT dir_fname, dir_lname, gen_title, COUNT(gen_title)
FROM director
-- Joining with 'movie_direction' using NATURAL JOIN to link directors with movies
NATURAL JOIN movie_direction
-- Further joining with 'movie_genres' using NATURAL JOIN to get additional movie genres information
NATURAL JOIN movie_genres
-- Joining with 'genres' using NATURAL JOIN to retrieve genre titles
NATURAL JOIN genres
-- Grouping the result by director names and genre titles
GROUP BY dir_fname, dir_lname, gen_title
-- Ordering the result by director first names and last names
ORDER BY dir_fname, dir_lname;
Sample Output:
dir_fname | dir_lname | gen_title | count
----------------------+----------------------+----------------------+-------
Alfred | Hitchcock | Mystery | 1
Bryan | Singer | Crime | 1
Danny | Boyle | Drama | 2
David | Lean | Adventure | 1
Frank | Darabont | Crime | 1
Hayao | Miyazaki | Animation | 1
Jack | Clayton | Horror | 1
James | Cameron | Action | 1
John | Boorman | Adventure | 1
Kevin | Spacey | Music | 1
Michael | Cimino | War | 1
Ridley | Scott | Thriller | 1
Sam | Mendes | Romance | 1
Stanley | Kubrick | Mystery | 1
Woody | Allen | Comedy | 1
(15 rows)
Code Explanation:
The said query in SQL that joins the tables director, movie_direction, movie_genres, and genres, and retrieves the first name and last name of the directors, the genre titles, and the count of movies directed by each director in each genre. The query groups the results by director first name, last name, and genre title, and sorts the results by director first name and last name.
The NATURAL JOIN keyword, which joins tables based on columns with the same name in each table. This joins the director table with the movie_direction table to get the movies directed by each director. Then joins the resulting table with the movie_genres table to get the genre for each movie and then it join the resulting table with the genres table to get the genre title for each genre ID.
The results then grouped by the director first name, last name, and genre title and then it sort the results by director first name and last name.
Alternative Solutions:
Using INNER JOIN and GROUP BY:
SELECT a.act_fname, a.act_lname, c.mov_title, c.mov_year
SELECT d.dir_fname, d.dir_lname, g.gen_title, COUNT(g.gen_title)
FROM director d
JOIN movie_direction md ON d.dir_id = md.dir_id
JOIN movie_genres mg ON md.mov_id = mg.mov_id
JOIN genres g ON mg.gen_id = g.gen_id
GROUP BY d.dir_fname, d.dir_lname, g.gen_title
ORDER BY d.dir_fname, d.dir_lname;
Explanation:
This query uses INNER JOINs to combine the director, movie_direction, movie_genres, and genres tables based on their respective IDs. It then applies a GROUP BY clause to group the results by director names (dir_fname, dir_lname) and genre titles (gen_title), along with a COUNT to get the count of each genre. The results are ordered by director names.
Using WHERE Clause with Table Aliases and GROUP BY:
SELECT a.act_fname, a.act_lname, c.mov_title, c.mov_year
SELECT d.dir_fname, d.dir_lname, g.gen_title, COUNT(g.gen_title)
FROM director d, movie_direction md, movie_genres mg, genres g
WHERE d.dir_id = md.dir_id
AND md.mov_id = mg.mov_id
AND mg.gen_id = g.gen_id
GROUP BY d.dir_fname, d.dir_lname, g.gen_title
ORDER BY d.dir_fname, d.dir_lname;
Explanation:
This query uses the older comma-separated syntax for joining tables and specifies the join conditions in the WHERE clause. It then applies a GROUP BY clause to group the results by director names and genre titles, along with a COUNT to get the count of each genre. The results are ordered by director names.
Relational Algebra Expression:
Relational Algebra Tree:
Practice Online
Query Visualization:
Duration:
Rows:
Cost:
Have another way to solve this solution? Contribute your code (and comments) through Disqus.
Previous: From the following tables, write a SQL query to find the actors who have not acted in any movie between1990 and 2000 (Begin and end values are included.). Return actor first name, last name, movie title and release year.
Next: From the following table, write a SQL query to find the movies with year and genres. Return movie title, movie year and generic title.
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
It will be nice if you may share this link in any developer community or anywhere else, from where other developers may find this content. Thanks.
https://w3resource.com/sql-exercises/movie-database-exercise/sql-exercise-movie-database-18.php
- Weekly Trends and Language Statistics
- Weekly Trends and Language Statistics