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@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@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@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!
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