To: pgsql-sql@postgresql.org
From: Thomas Kellerer <spam_eater@gmx.net>
Subject: Re: left outer join only select newest record
Date: Wed, 23 May 2012 12:22:56 +0200
Lines: 39
Message-ID: <jpidmq$dec$1@dough.gmane.org>
References: <201205231027.39198.gary.stainburn@ringways.co.uk>
	<53C6FD0FE48B43C0AD419648FDC8D943@marktestcr.marktest.pt>
	<201205231047.27372.gary.stainburn@ringways.co.uk>
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In-Reply-To: <201205231047.27372.gary.stainburn@ringways.co.uk>

Gary Stainburn, 23.05.2012 11:47:
> Here is a select to show the problem. There is one stock record and two tax
> records. What I'm looking for is how I can return only the second tax record,
> the one with the highest ud_id
>
> select s_stock_no, s_regno, s_vin, s_created, ud_id, ud_handover_date from
> stock s left outer join used_diary u on s.s_regno = u.ud_pex_registration
> where s_stock_no = 'UL15470';
>
>   s_stock_no | s_regno |       s_vin       |         s_created          |
> ud_id | ud_handover_date
> ------------+---------+-------------------+----------------------------+-------+------------------
>   UL15470    | YG12*** | KNADN312LC6****** | 2012-05-21 09:15:31.569471 |
> 41892 | 2012-04-06
>   UL15470    | YG12*** | KNADN312LC6****** | 2012-05-21 09:15:31.569471 |
> 42363 | 2012-05-16
> (2 rows)
>

Something like:

select *
from (
     select s_stock_no,
            s_regno
            s_vin,
            s_created,
            ud_id,
            ud_handover_date,
            row_number() over (partition by s_stock_no order by ud_id desc) as rn
     from stock s
       left outer join used_diary u on s.s_regno = u.ud_pex_registration
     where s_stock_no = 'UL15470'
) t
where rn = 1


The "partition by s_stock_no order" isn't really necessary as your where clause already limits that to a single stock_no.
But in case you change that statement to return more than one stock_no in the future it will be necessary.