Re: How to reference a subquery column alias?

Hello Sergey

Thanks by your answer, this worked after i add at the end an AS xxx
clause, because an error telling me something about subqueries in FROM
must have an alias, so i did it:

SELECT *, min_caudal + max_caudal AS diferencia FROM (
...your expression...
) AS temp

and worked

Best,

Sergey Konoplev escribió:
> Hi,
>
> Just wrap your expression with another SELECT and operate with the aliases like
>
> SELECT *, min_caudal + max_caudal AS diferencia FROM (
> ...your expression...
> )
>
> 2010/8/9 José María Terry Jiménez <jtj@tssystems.net>:
>
>> Hello
>>
>> Yesterday a list user solved me a problem with a sententence with two
>> subqueries. The solution was this:
>>
>> SELECT remota_id,min(hora),max(hora), (SELECT caudal FROM historicos WHERE
>> remota_id=ooo.remota_id AND hora=min(ooo.hora)) as min_caudal, (SELECT
>> caudal FROM historicos WHERE remota_id=ooo.remota_id AND hora=max(ooo.hora))
>> as max_caudal from historicos AS ooo GROUP BY remota_id ORDER BY remota_id;
>>
>> The issue now is i want to do some calculations with the subqueries columns
>> (min_caudal and max_caudal), for example adding them. (Get a new column with
>> max_caudal and min_caudal (alias) added)
>>
>> I have tried to add in the SELECT;
>> ,max_caudal+min_caudal as diferencia
>> ,ooo.max_caudal+ooo.min_caudal as diferencia
>> ,historicos.max_caudal+historicos.min_caudal as diferencia
>> ,(SELECT max_caudal+min_caudal) as diferencia
>>
>> I've read the SELECT and Table Expressions documentation pages, but didn't
>> found a solution.
>>
>> Can anyone tell me how to reference or make the calculation with those alias
>> names?
>>
>> TIA
>>
>> Best,
>>
>>

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
Scanned with Copfilter Version 0.84beta3a (ProxSMTP 1.6)
AntiVirus: ClamAV 0.95.2/11523 - Mon Aug  9 19:20:40 2010
by Markus Madlener @ http://www.copfilter.org