| From: | negora <negora(at)negora(dot)com> |
|---|---|
| To: | Jayadevan M <Jayadevan(dot)Maymala(at)ibsplc(dot)com> |
| Cc: | pgsql-sql(at)postgresql(dot)org |
| Subject: | Re: SUM the result of a subquery. |
| Date: | 2010-09-02 12:36:00 |
| Message-ID: | 4C7F9A30.2010908@negora.com |
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| Lists: | pgsql-sql |
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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(dot)CC04F696-ON65257792(dot)00446E7D-65257792(dot)00448A26(at)ibsplc(dot)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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</pre>
</blockquote>
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