OK, well that was painful but we are stronger for the effort, thanks
for bringing up the issue. r5740 of trunk will allow your original
mapper(A.join(B))->mapper(B) to configure properly.
On Jan 28, 2009, at 11:28 PM, Michael Bayer wrote:
>
>
> a join is of the form:
>
> table1.join(table2, onclause)
>
> such as
>
> subscriber_table.join(address_table,
> and_(address_table.c.subscriber_id==subscriber.c.id,
> address_table.c.type=='MAIN'))
>
> but unfortunately current relation() code does not support a join of
> X/
> Y to Y, unless the join of X/Y is assembled via joined table
> inheritance. As a workaround, you can wrap your join() in an
> aliased select(). A fix may be available in the next 10 minutes or
> maybe not.
>
> You also could forego the complexity of mapping to a join and just
> modify your Subscriber class to break up the "addresses" collection
> amongst a proxy of the "MAIN" element and a list of the remaining
> elements. an attribute_mapped_collection could help to accomplish
> this nicely.
>
>
>
> On Jan 28, 2009, at 7:22 PM, GHZ wrote:
>
>>
>> I have a subscriber and address table.
>>
>> a subscriber will have one and only one 'MAIN' address.
>> I want the subscriber and MAIN address to be represented by one class
>> 'Subscriber'. However, I want that class to have a collection
>> 'addresses' which contains other addresses (e.g. old addresses) - (it
>> can include the 'MAIN' address too .. or not.. I don't care)
>>
>> subscriber_table = Table('subscriber', metadata,
>> Column('id', primary_key=True),
>> autoload=True)
>>
>> address_table = Table('address',
>> metadata,
>> Column('subscriber_id', ForeignKey
>> ('subscriber.id'), primary_key=True),
>> Column('address_type', primary_key=True),
>> autoload=True)
>>
>>
>>
>> subscriber_with_default_address = sql.join( subscriber_table.c.id
>> == address_table.c.subscriber_id).??? <- something to say
>> address_table.type is 'MAIN'
>>
>> mapper(Address, address_table)
>>
>> mapper(Subscriber, subscriber_and_address, properties={
>> 'id':[subscriber_table.c.id, address_table.c.subscriber_id],
>> 'addresses' : relation(Address, collection_class=Addresses,
>> backref='customer')
>> })
>>
>> a) I can't quite figure out how to say (address.type is default)
>> b) even without this I get:
>>
>> sqlalchemy.exc.ArgumentError: Can't determine relation direction for
>> relationshi
>> p 'Subscriber.addresses' - foreign key columns are present in both
>> the
>> parent an
>> d the child's mapped tables. Specify 'foreign_keys' argument.
>>
>> if I do specify foreign_keys parameter to the relation function, then
>> I still get the same.
>>
>> Thanks
>>
>>
>>>
>
>
> >
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