SELECT CASE, COUNT(*) - mysql

Question

I want to select the number of users that has marked some content as favorite and also return if the current user has "voted" or not. My table looks like this
CREATE TABLE IF NOT EXISTS `favorites` (
`user` int(11) NOT NULL DEFAULT '0',
`content` int(11) NOT NULL DEFAULT '0',
PRIMARY KEY (`user`,`content`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 ;
Say I have 3 rows containing
INSERT INTO `favorites` (`user`, `content`) VALUES
(11, 26977),
(22, 26977),
(33, 26977);
Using this
SELECT COUNT(*), CASE
WHEN user='22'
THEN 1
ELSE 0
END as has_voted
FROM favorites WHERE content = '26977'
I expect to get has_voted=1 and COUNT(*)=3 but
I get has_voted=0 and COUNT(*)=3. Why is that? How to fix it?

Answer 1

This is because you mixed aggregated and non-aggregated expressions in a single SELECT. Aggregated expressions work on many rows; non-aggregated expressions work on a single row. An aggregated (i.e. COUNT(*)) and a non-aggregated (i.e. CASE) expressions should appear in the same SELECT when you have a GROUP BY, which does not make sense in your situation.
You can fix your query by aggregating the second expression - i.e. adding a SUM around it, like this:
SELECT
COUNT(*) AS FavoriteCount
, SUM(CASE WHEN user=22 THEN 1 ELSE 0 END) as has_voted
FROM favorites
WHERE content = 26977
Now both expressions are aggregated, so you should get the expected results.

Answer 2

Try this with SUM() and without CASE
SELECT
COUNT(*),
SUM(USER = '22') AS has_voted
FROM
favorites
WHERE content = '26977'
See Fiddle Demo

Answer 3

Try this:
SELECT COUNT(*), MAX(USER=22) AS has_voted
FROM favorites
WHERE content = 26977;
Check the SQL FIDDLE DEMO
OUTPUT
| COUNT(*) | HAS_VOTED |
|----------|-----------|
| 3 | 1 |

Answer 4

You need sum of votes.
SELECT COUNT(*), SUM(CASE
WHEN user='22'
THEN 1
ELSE 0
END) as has_voted
FROM favorites WHERE content = '26977'

Answer 5

You are inadvertently using a MySQL feature here: You aggregate your results to get only one result record showing the number of matches (aggregate function COUNT). But you also show the user (or rather an expression built on it) in your result line (without any aggregate function). So the question is: Which user? Another dbms would have given you an error, asking you to either state the user in a GROUP BY or aggregate users. MySQL instead picks a random user.
What you want to do here is aggregate users (or rather have your expression aggregated). Use SUM to sum all votes the user has given on the requested content:
SELECT
COUNT(*),
SUM(CASE WHEN user='22' THEN 1 ELSE 0 END) as sum_votes
FROM favorites
WHERE content = '26977';

Answer 6

You forgot to wrap the CASE statement inside an aggregate function. In this case has_voted will contain unexpected results since you are actually doing a "partial group by". Here is what you need to do:
SELECT COUNT(*), SUM(CASE WHEN USER = 22 THEN 1 ELSE 0 END) AS has_voted
FROM favorites
WHERE content = 26977
Or:
SELECT COUNT(*), COUNT(CASE WHEN USER = 22 THEN 1 ELSE NULL END) AS has_voted
FROM favorites
WHERE content = 26977