A-Original-Client-UID: 188.129.106.69 splivalo
Message-ID: <4CFBC24A.90708@megafon.hr>
Date: Sun, 05 Dec 2010 17:48:10 +0100
From: Mario Splivalo <mario.splivalo@megafon.hr>
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To: pgsql-sql@postgresql.org
Subject: Re: Get the max viewd product_id for user_id
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On 12/03/2010 12:40 PM, Jayadevan M wrote:
> Hello,
>
>> I went this way, but for a large number of user_id's, it's quite slow:
>>
>> CREATE VIEW v_views AS
>>     SELECT user_id, product_id, count(*) as views
>>     FROM viewlog
>>     GROUP BY user_id, product_id
>>
>> SELECT
>>     DISTINCT user_id,
>>     (SELECT product_id FROM v_views inn WHERE inn.user_id = out.user_id
>> ORDER BY views DESC LIMIT 1) as product_id,
>>     (SELECT views FROM v_views inn WHERE inn.user_id = out.user_id ORDER
> BY
>> views DESC LIMIT 1) as views
>> FROM
>>     v_views out
>>
> Does this work faster?
> select x.user_id,y.product_id,x.count from
> (select user_id, max(count ) as count from (select user_id,product_id,
> count(*) as count from viewlog group by user_id,product_id) as x group by
> user_id
> ) as x inner join
> (select user_id,product_id, count(*) as count1 from viewlog group by
> user_id,product_id ) as y
> on x.user_id=y.user_id and x.count=y.count1
>

It does, yes. Actually, pretty silly of me not to implement it that way, 
thank you.

Since I already have the view, the query now looks like this:


select
	x.user_id,
	y.product_id,
	x.views
from (
	select
		user_id,
		max(views) as views
	from
		v_views
	group by
		user_id
) as x
	inner join v_views as y
	on x.user_id=y.user_id and x.views=y.views

And CTEs would also help here :)

	Mario