Re: Find users that have ALL categories

On Wed, Jun 30, 2010 at 12:11:35AM -0700, Nick wrote:
> Is this the most efficient way to write this query? Id like to get a
> list of users that have the categories 1, 2, and 3?
>
> SELECT user_id FROM user_categories WHERE category_id IN (1,2,3) GROUP
> BY user_id HAVING COUNT(*) = 3
>
> users_categories (user_id, category_id)
> 1 | 1
> 1 | 2
> 1 | 3
> 2 | 1
> 2 | 2
> 3 | 1
> 4 | 1
> 4 | 2
> 4 | 3
>
> The result should produce 1 & 4.

The above method depends on (user_id, category_id) being unique, and
excludes users with, say, categories 1, 2, 3 and 4. Are you sure that
that latter is what you want?

This is, I believe, a little clearer as to what it's actually doing,
and doesn't exclude user_ids with more matches:

SELECT user_id
FROM user_categories
GROUP BY user_id
HAVING array_agg(category_id) @> ARRAY[1,2,3]
ORDER BY user_id; /* Not really needed, but could be handy */

In 9.0, you'll be able to use the following to get only exact matches:

SELECT user_id
FROM user_categories
GROUP BY user_id
HAVING array_agg(category_id ORDER BY category_id) = ARRAY[1,2,3]
ORDER BY user_id; /* Not really needed, but could be handy */

Until then, you can make an array_sort() function like this:

CREATE OR REPLACE FUNCTION array_sort(ANYARRAY)
RETURNS ANYARRAY
LANGUAGE SQL
STRICT
AS $$
SELECT ARRAY(
SELECT unnest($1) AS i
ORDER BY i
);
$$;

then use it like this:

SELECT user_id
FROM user_categories
GROUP BY user_id
HAVING array_sort(array_agg(category_id)) = ARRAY[1,2,3]
ORDER BY user_id;

to get only exact matches.

As to speed, you'd have to test on your actual data sets. Indexing
user_id may help here.

Cheers,
David.
--
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