On Mon, Mar 12, 2012 at 16:18:05 -0400,
Michael Gould mgo...@isstrucksoftware.net wrote:
You need to include all columns that are not aggregrative columns in the group
by. Even though that is the standard it is a pain to list all columns even if
you don't need them
In later versions of
Thanks that is a help. I would be nice if any key could be used as those are
normally the things I would do group by's
Regards
Mike Gould
From my Samsung Android tablet on T-Mobile. The first nationwide 4G
networkBruno Wolff III br...@wolff.to wrote:On Mon, Mar 12, 2012 at 16:18:05
-0400,
On Mon, Apr 09, 2012 at 13:55:04 -0400,
Michael Gould mgo...@isstrucksoftware.net wrote:
Thanks that is a help. I would be nice if any key could be used as those are
normally the things I would do group by's
This is what the 9.1 documentation says:
When GROUP BY is present, it is not valid
You need to include all columns that are not aggregrative columns in the group
by. Even though that is the standard it is a pain to list all columns even if
you don't need them
Best Regards
Michael Gould
Sent from Samsung mobile
Alexander Reichstadt l...@mac.com wrote:
Hi,
the following
Hi,
the following statement worked on mysql but gives me an error on postgres:
column addresses.address1 must appear in the GROUP BY clause or be used in an
aggregate function
I guess I am doing something wrong. I read the web answers, but none of them
seem to meet my needs:
SELECT
Hi,
You can use one of windowing function:
http://www.postgresql.org/docs/9.1/static/tutorial-window.html
http://www.postgresql.org/docs/9.1/static/functions-window.html
this could be rank() in subquery or first_value(vale any), but there could
be performance issue
another solution could be
So the mysql way for group by seems to be non-standard.
What works for postgres is the DISTINCT ON (fieldname) approach.
Thanks
Am 12.03.2012 um 20:35 schrieb Alexander Reichstadt:
Hi,
the following statement worked on mysql but gives me an error on postgres:
column addresses.address1
Instead of the joins you can use a subquery to get the first address.
Or you can do the joins without the group by and use row_number()
over(partition by companies.id) on the select list to label each company
address with a number starting at 1. You can just keep rows that have
row_number =
Thanks, I just posted my response to my own question for the archives. I take
it also that group by is faster than distinct on. If it is a substantial
performance gain I have to work on this some more. A subquery I would expect
would be much of a drag, so for all keystroke-updated list-tables
On Mon, Mar 12, 2012 at 1:35 PM, Alexander Reichstadt l...@mac.com wrote:
Hi,
the following statement worked on mysql but gives me an error on postgres:
column addresses.address1 must appear in the GROUP BY clause or be used in
an aggregate function
I guess I am doing something wrong. I
I guess I lack the knowledge to integrate your answer in my queryActually
I'd prefer to always see the first address entered unless there is a
where-clause added. Not sure how this works out then and haven't tested. But
given the initial query extended by distinct on it would be like so:
On Mon, Mar 12, 2012 at 3:19 PM, Alexander Reichstadt l...@mac.com wrote:
But where would I insert the max(address) piece?
Just put max() or min() around any field in the select list that's not
in the group by clause
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