Weird MySQL SELECT WHERE IN problem

Question

I'm trying to do this:

SELECT * FROM library WHERE $cat IN (tags);

Say that I have $cat = '15'; and tags is a fieldname in library that looks like this: 2,15,34, 45 How can I get it to work? I understand that normally the $var should be after the IN ($var), but the imploded array sits in my DB and not in the $var - somehow I've gotten it the wrong way around and I just cannot figure it out - Please help me, I'm wasting days :)

Consider to normalize your table instead of using find_in_set() — Nicola Cossu, Mar 17 '11 at 15:23
See [this question](http://stackoverflow.com/questions/4155873/mysql-find-in-set-vs-in). `IN` does not work this way. — gen_Eric, Mar 17 '11 at 15:26

score 1 · Accepted Answer · answered Mar 17 '11 at 15:21

1

You can use FIND_IN_SET instead, like this:

SELECT * FROM library
WHERE FIND_IN_SET($cat, tags);

answered Mar 17 '11 at 15:21

a'r

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score 1 · Answer 2 · answered Mar 17 '11 at 15:22

1

select * from library where tags like '%,$cat,%';

answered Mar 17 '11 at 15:22

Beofett

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This doesn't work when the required tag is at the start or the end of the list. – a'r Mar 17 '11 at 16:41
You can use this trick to get around it: `CONCAT(',',tags,',') LIKE '%,$cat,%'`, but in MySQL, `FIND_IN_SET` is much easier. – a'r Mar 17 '11 at 16:55

score 0 · Answer 3 · answered Mar 17 '11 at 15:24

0

Would you not do:

SELECT * FROM library WHERE tags = $cat

answered Mar 17 '11 at 15:24

anothershrubery

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Weird MySQL SELECT WHERE IN problem

3 Answers3