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.
--
David Fetter <david@fetter.org> http://fetter.org/
Phone: +1 415 235 3778 AIM: dfetter666 Yahoo!: dfetter
Skype: davidfetter XMPP: david.fetter@gmail.com
iCal: webcal://www.tripit.com/feed/ical/people/david74/tripit.ics

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On Thu, Jul 01, 2010 at 04:26:38AM -0700, David Fetter wrote:

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

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?

AFAICT, the above code will include a user with categories 1 to 4. Why
do you think otherwise?

If the (user_id,category_id) combination isn't unique, it's easy to
change the HAVING clause into HAVING COUNT(DISTINCT category_id) = 3.

--
Sam http://samason.me.uk/

On Thu, Jul 01, 2010 at 12:37:55PM +0100, Sam Mason wrote:

On Thu, Jul 01, 2010 at 04:26:38AM -0700, David Fetter wrote:

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

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?

AFAICT, the above code will include a user with categories 1 to 4.
Why do you think otherwise?

If the (user_id,category_id) combination isn't unique, it's easy to
change the HAVING clause into HAVING COUNT(DISTINCT category_id) =
3.

Oops. You're right, of course. That's what I get for posting before
waking up. ;)

Cheers,
David.
--
David Fetter <david@fetter.org> http://fetter.org/
Phone: +1 415 235 3778 AIM: dfetter666 Yahoo!: dfetter
Skype: davidfetter XMPP: david.fetter@gmail.com
iCal: webcal://www.tripit.com/feed/ical/people/david74/tripit.ics

Remember to vote!
Consider donating to Postgres: http://www.postgresql.org/about/donate