Query not returning the expected results - mysql

Question

SELECT u.* ,
(select CASE u.ID
WHEN u.ID in (select RequestedUserID from user_requestes where userID=3) THEN 0
ELSE 1
END ) AS Accepted
FROM users u
WHERE u.ID <>3
and u.id not in (select friends.FriendID
from friends
where friends.UserID=3 or friends.FriendID=3)
order by u.Name asc
i am trying to execute this query using phpmyadmin
select RequestedUserID from user_requestes where userID=3
the above query return 79 as result
and if i execute the original query i found this
Accepted should be 0 and not 1

Answer 1

And what if you write your query like this?
SELECT
u.*,
CASE
WHEN u.ID IN (select RequestedUserID from user_requestes where userID=3) THEN 0
ELSE 1
END AS Accepted
FROM
users u
WHERE
u.ID <>3
and u.id not in (
select
friends.FriendID
from
friends
where
friends.UserID=3 or friends.FriendID=3
)
order by
u.Name asc
Don't use SELECT before CASE;
If you go for CASE WHEN... syntax, you must provide values and not search conditions (See MySQL documentation here)

Answer 2

There is a mistake in your case expression
CASE u.ID WHEN u.ID in (...)
reads as follows: look up u.id in the subquery. Found = true, not found = false. In MySQL true = 1 and false = 0.
CASE u.ID WHEN <either 1 or 0>
You are mistakenly comparing the user ID with the boolean result 1 or 0.
You want this instead:
SELECT
u.* ,
CASE
WHEN u.ID in (select RequestedUserID from user_requestes where userID=3) THEN 0
ELSE 1
END AS Accepted
FROM ...
By the way: There is probably a semantical mistake in your friends subquery, as it is always FriendID you are returning. I suppose that should be:
and u.id not in
(
select case when FriendID = 3 then UserID else FriendID end
from friends
where UserID = 3 or FriendID = 3
)
or simply
and u.id not in (select FriendID from friends where UserID = 3)
and u.id not in (select UserID from friends where FriendID = 3)

Answer 3

This might work better:
SELECT u.*,
IF(EXISTS (
SELECT *
from user_requestes
where userID = 3
AND RequestedUserID = u.ID ),
0, 1 ) AS Accepted
FROM users AS u
LEFT JOIN friends AS f ON f.FriendID = u.ID
WHERE u.ID != 3
AND f.FriendID IS NULL
AND ( f.UserID = 3 or f.FriendID = 3 )
Note the use of EXISTS as being more efficient than IN ( SELECT ... ). Ditto for LEFT JOIN ... IS NULL.