This seems very basic but I can't figure it out.
I've got a table "item_tags", and I want to select all of the items that match tags 1 and 2 (as in, each item has to have both tags).
How would I do this 开发者_开发问答in mysql?
Create table is:
CREATE TABLE `item_tags` (
`uid_local` int(11) NOT NULL DEFAULT '0',
`uid_foreign` int(11) NOT NULL DEFAULT '0',
`sorting` int(11) NOT NULL DEFAULT '0',
KEY `uid_local` (`uid_local`),
KEY `uid_foreign` (`uid_foreign`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8;
Thanks!
Use:
SELECT i.uid
FROM ITEMS i
JOIN ITEM_TAGS it ON it.uid_local = i.uid
AND it.uid_foreign IN (1, 2)
GROUP BY i.uid
HAVING COUNT(DISTINCT it.uid_foreign) = 2
You need to have a GROUP BY and HAVING clause defined, and the count of distinct tag ids must equal the number of tags you specify in the IN clause.
something like this?
SELECT i.* from items i inner join items_tags it
on i.id = it.item_id
inner join tags t
on t.id = it.tag_id
WHERE t.name in ('tag1', 'tag2');
EDIT:
suppouse you have items_tags: (item_id, tag_id) as table
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