A-Original-Client-UID: 188.129.106.69 splivalo
Message-ID: <4CFBC46D.1080105@megafon.hr>
Date: Sun, 05 Dec 2010 17:57:17 +0100
From: Mario Splivalo <mario.splivalo@megafon.hr>
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To: Jayadevan M <Jayadevan.Maymala@ibsplc.com>
CC: pgsql-sql@postgresql.org
Subject: Re: Get the max viewd product_id for user_id
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	<OF16FA4770.DCF7942F-ON652577EE.003F9FF3-652577EE.0040227D@ibsplc.com>
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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

The issue in both approaches is that if I have two product_ids that are 
viewed same number of times and share the first place as most viewed 
products by that user, I'll get only one of them (LIMIT 1 OR MAX() can 
only return one row :).

I don't see how I can sort this out with elegance in SQL.

	Mario