From: negora <negora@negora.com>
To: Jayadevan M <Jayadevan.Maymala@ibsplc.com>
Cc: pgsql-sql@postgresql.org
Subject: Re: SUM the result of a subquery.
Date: Thu, 02 Sep 2010 14:36:00 +0200
Message-ID: <4C7F9A30.2010908@negora.com> (raw)
In-Reply-To: <OF39AE7609.CC04F696-ON65257792.00446E7D-65257792.00448A26@ibsplc.com>
References: <4C7F95DE.9020200@negora.com>
<OF39AE7609.CC04F696-ON65257792.00446E7D-65257792.00448A26@ibsplc.com>
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<font face="Verdana">Wow, I had no idea about this kind of SELECT
expression. It works flawless!!! Thank you lots Jayadevan</font><font
face="Verdana"> :) .</font><br>
<font face="Verdana"><br>
</font><br>
On 02/09/10 14:28, Jayadevan M wrote:
<blockquote
cite="mid:OF39AE7609.CC04F696-ON65257792.00446E7D-65257792.00448A26@ibsplc.com"
type="cite">
<blockquote type="cite">
<pre wrap="">SELECT SUM (
(SELECT i.id_item, i.price, SUM (o.quantity), ROUND (SUM
(o.quantity) * i.price, 2) AS cost
FROM orders o
JOIN items i ON i.id_item = o.id_item
WHERE o.date_order BETWEEN '2010-01-01' AND '2010-01-31'
GROUP BY i.id_item, i.price)
);
No luck. Obviously SUM expects an expression, not a set of rows. Is
there a way to perform a sum of the resulting rows?
</pre>
</blockquote>
<pre wrap="">I don't have a PostgreSQL server to try this right now. But you are
looking for something like
SELECT SUM (cost) from (
(SELECT i.id_item, i.price, SUM (o.quantity), ROUND (SUM (o.quantity) *
i.price, 2) AS cost
FROM orders o
JOIN items i ON i.id_item = o.id_item
WHERE o.date_order BETWEEN '2010-01-01' AND '2010-01-31'
GROUP BY i.id_item, i.price)
) as x
Regards,
Jayadevan
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</blockquote>
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To: pgsql-sql@postgresql.org
Cc: negora@negora.com, Jayadevan.Maymala@ibsplc.com
Subject: Re: SUM the result of a subquery.
In-Reply-To: <4C7F9A30.2010908@negora.com>
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