SQL exercises on movie Database: Find the names of all reviewers who have ratings with a NULL value
SQL movie Database: Subquery Exercise-8 with Solution
8. From the following table, write a SQL query to find those reviewers who have not given a rating to certain films. Return reviewer name.
Sample table: reviewerrev_id | rev_name --------+-------------------------------- 9001 | Righty Sock 9002 | Jack Malvern 9003 | Flagrant Baronessa 9004 | Alec Shaw 9005 | 9006 | Victor Woeltjen 9007 | Simon Wright 9008 | Neal Wruck 9009 | Paul Monks 9010 | Mike Salvati 9011 | 9012 | Wesley S. Walker 9013 | Sasha Goldshtein 9014 | Josh Cates 9015 | Krug Stillo 9016 | Scott LeBrun 9017 | Hannah Steele 9018 | Vincent Cadena 9019 | Brandt Sponseller 9020 | Richard AdamsSample table: rating
mov_id | rev_id | rev_stars | num_o_ratings --------+--------+-----------+--------------- 901 | 9001 | 8.40 | 263575 902 | 9002 | 7.90 | 20207 903 | 9003 | 8.30 | 202778 906 | 9005 | 8.20 | 484746 924 | 9006 | 7.30 | 908 | 9007 | 8.60 | 779489 909 | 9008 | | 227235 910 | 9009 | 3.00 | 195961 911 | 9010 | 8.10 | 203875 912 | 9011 | 8.40 | 914 | 9013 | 7.00 | 862618 915 | 9001 | 7.70 | 830095 916 | 9014 | 4.00 | 642132 925 | 9015 | 7.70 | 81328 918 | 9016 | | 580301 920 | 9017 | 8.10 | 609451 921 | 9018 | 8.00 | 667758 922 | 9019 | 8.40 | 511613 923 | 9020 | 6.70 | 13091
Sample Solution:
-- Selecting distinct reviewer names from the 'reviewer' table
-- Using a subquery to find rev_id from the 'rating' table based on a condition
-- Specifically, finding rev_id where rev_stars is NULL
SELECT DISTINCT rev_name
FROM reviewer
WHERE rev_id IN (
-- Subquery to find rev_id with rev_stars being NULL
SELECT rev_id
FROM rating
WHERE rev_stars IS NULL
);
Sample Output:
rev_name -------------------------------- Neal Wruck Scott LeBrun (2 rows)
Code Explanation:
The said query in SQL which selects distinct reviewer names from the 'reviewer' table where the corresponding reviewer IDs have at least one review with a null rating in the 'Rating' table. The results are ordered by reviewer name in ascending order.
The query uses a subquery to find the reviewer IDs with null ratings, and then selects the distinct reviewer names corresponding to those IDs. The "DISTINCT" keyword ensures that only unique reviewer names are returned.
Alternative Solutions:
Using INNER JOIN:
SELECT DISTINCT r.rev_name
FROM reviewer r
JOIN rating ra ON r.rev_id = ra.rev_id
WHERE ra.rev_stars IS NULL;
Explanation:
This solution uses an INNER JOIN to combine the reviewer and rating tables based on rev_id. It then filters the result to only include reviewers whose rev_stars are NULL.
Using EXISTS:
SELECT DISTINCT r.rev_name
FROM reviewer r
WHERE EXISTS (
SELECT 1
FROM rating ra
WHERE ra.rev_id = r.rev_id AND ra.rev_stars IS NULL
);
Explanation:
This query utilizes the EXISTS clause to check if there exists a record in the rating table with a NULL rev_stars for each reviewer in the reviewer table. If such a record exists, the reviewer's rev_name is included in the result set.
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 table, write a SQL query to find those movies, which have no ratings. Return movie title.
Next: From the following tables, write a SQL query to find those movies, which reviewed by a reviewer and got a rating. Sort the result-set in ascending order by reviewer name, movie title, review Stars. Return reviewer name, movie title, review Stars.
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-29.php
- Weekly Trends and Language Statistics
- Weekly Trends and Language Statistics