MySQL COUNT results of UNION ALL statement

问题

I'm sure there must be a way to do this but my MySQL knowledge is holding me back.

I have a single table that stores page tags

page_id          tag
51               New Zealand
51               Trekking
58               UK
77               New Zealand
77               Trekking
89               City Break
101              Shopping
...

I want to do a search for pages that have two tags, e.g. "New Zealand" and "Trekking". I've looked at UNIONS, INTERSECT (equiv), JOINS and I can't work out what is the best way to do it. The best I have come up with is to do:

SELECT page_id FROM tags
WHERE tag = "New Zealand"
UNION ALL
SELECT page_id FROM tags
WHERE tag = "Trekking"
... and then some kind of COUNT on those pages that feature twice??

I essentially want to UNION the two searches together and 'keep' the duplicates and discard the rest. Is this possible in a simple way or do I need to start getting complex with it?

回答1:

If I understood you correctly, this should do it:

SELECT page_id, count(*)
FROM tags
WHERE tag IN ('New Zealand', 'Trekking')
GROUP BY page_id
HAVING count(*) > 1

You don't need to use a UNION if you select from the same table.

回答2:

Considering that you want page_id that have both tag "New Zealand" and "Trekking"


SELECT t1.page_id
FROM tags t1, tags t2
WHERE t1.page_id = t2.page_id
AND
t1.tag="New Zealand"
AND
t2.tag="Trekking"

Seems to be Right, hmm but check once..., will ping back if i found easier and more accurate

回答3:

try this

CREATE TABLE temporary
SELECT page_id FROM tags
WHERE tag = "New Zealand"
UNION ALL
SELECT page_id FROM tags
WHERE tag = "Trekking"

then try this

SELECT COUNT(*) FROM temporary

don't forget to

DROP TABLE temporary

来源：https://stackoverflow.com/questions/8111440/mysql-count-results-of-union-all-statement

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mysql

select

union