I think it might not exist b/c it isn't very SQLObjecty.
You'd be returning an object with a new field in it, SUM(), which isn't
part of the SQLObject and probably a subset of the SQLObject fields as
well. This isn't very OO which is what I think SQLObject was designed for.
On Fri, 4 Aug 2006, Sean O'Donnell wrote:
> Yip, us poor seans seem to be flat out of luck. Im off to a wedding for
> the weekend, when I get back
> I will see about getting my hands dirty. I've never looked at the
> SQLObject source, so it should prove
> to be an education.
>
> Thanks for getting back to me so quickly.
>
> Sean
>
> Sean McBride wrote:
>> Unfortunately, it's not implemented right now, and it seems that only
>> Seans need it! See my post a little ways down this list about it. As it
>> currently stands, someone like you or I is going to have to implement it
>> and submit to the list before this kind of ordering is possible.
>> (Believe me, I have about 5-8 queries right now that could really use it.)
>>
>> I'm considering trying my hand at adding it but won't have time for at
>> least a few more weeks due to my current project.
>>
>> - Sean
>>
>> Sean O'Donnell wrote:
>>
>>> Hi There,
>>>
>>> Im trying to figure out a way to order a query by the sum of a colum in
>>> a MultipleJoin
>>>
>>> e.g (not the original code, its messed up by me trying so many possible
>>> solutions :)
>>>
>>> class Salesman(SQLObject):
>>> name = StringCol()
>>> sales = MultipleJoin(Sale)
>>>
>>> class Sale(SQLObject):
>>> salesman = ForeignKey('Salesman')
>>> amount = IntCol()
>>>
>>>
>>> Now supposing I want to query a list of all of the Salesmen, ordered by
>>> the Sum of the amounts of their sales?
>>> Will SQLObject let me do that?
>>>
>>> If I was writing plain ole SQL I would do
>>>
>>> select salesman.name, sum(sale.amount) from salesman, sale where
>>> salesman.id = sale.salesman_id group by salesman.id order by sum(sale)
>>> desc.
>>>
>>> I dug through the SQLObject documentation and order by only seems to
>>> take single column names, I found some references to group by on this
>>> mailing list, but could not get anything working :(
>>>
>>> Thanks for your help
>>>
>>> Sean
>>>
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