[RndTbl] SQL problem

[John Lange]

Sorry, I see that was totally wrong... You'd need something like below but
I don't think this will work as-is...

SELECT custid as cid, (select count(*) FROM ordertable where custid=cid and
TTL=X), (select count(*) FROM ordertable where custid=cid and TTL=Y)
FROM ordertable
order by custid;

On Thu, Mar 24, 2016 at 11:50 AM, John Lange <john at johnlange.ca> wrote:

> What Adam said; something like:
>
> SELECT custid, (select count(*) FROM ordertable where TTL=X), (select
> count(*) FROM ordertable where TTL=Y)
> FROM ordertable
> order by custid;
>
>
> On Thu, Mar 24, 2016 at 9:21 AM, Adam Thompson <athompso at athompso.net>
> wrote:
>
>> Subqueries:
>> SELECT (SELECT ...), (SELECT ...) FROM ...
>>
>> If they don't work, upgrade to a newer version of MariaDB. (If this is
>> the customer I assume it is, use RackSpace's IUS repo to upgrade).
>>
>> Alternately, create two VIEWs, one for each aggregate count, and OUTER
>> JOIN them.
>> -Adam
>>
>>
>> On March 24, 2016 9:16:31 AM CDT, Trevor Cordes <trevor at tecnopolis.ca>
>> wrote:
>>>
>>> Can anyone help me figure out how to do this in SQL (MySQL)?
>>>
>>> (pseudo-code giving you the gist):
>>>
>>> SELECT custid, (count orders where ttl=X), (count orders where ttl=Y)
>>> FROM ordertable
>>> GROUP BY custid;
>>>
>>>
>>> I basically want to count two different things based on two different
>>> wheres.  If I put in a where clause then I select either the X or the Y
>>> but I can't seem to get both in such a way I can count the X's and Y's.  I
>>> tried thinking about unions but couldn't make it work.  I want to do this
>>> all in 1 query as I want the sorted union of all custid's (X and Y).
>>>
>>> P.S. ordertable has/can have multiple entries for each custid.
>>>
>>> Ideas are appreciated!
>>> ------------------------------
>>>
>>> Roundtable mailing list
>>> Roundtable at muug.mb.ca
>>> http://www.muug.mb.ca/mailman/listinfo/roundtable
>>>
>>>
>> --
>> Sent from my Android device with K-9 Mail. Please excuse my brevity.
>>
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>>
>
>
> --
> John Lange
> www.johnlange.ca
>



-- 
John Lange
www.johnlange.ca
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