> On Oct 21, 2014, at 6:07 PM, Jonathan Vanasco <[email protected]> wrote:
>
> I've been staring at this for a while, and can't figure out a way to make the
> mapper happy:
>
> i have 3 Classes (tables):
>
> * List (list)
> * ListItem (list_item)
> * ItemType1 (item_type_1)
> * ItemType2 (item_type_2)
> * ItemType3 (item_type_3)
>
> until now i've been using a joinedload
>
> query = s.query(List)\
> .options(
> joinedload('list_item'),
> joinedload('list_item.item_type_1'),
> joinedload('list_item.item_type_2'),
> joinedload('list_item.item_type_3'),
> )
>
> The performance is starting to become less than optimal, so I wanted to try
> creating a subquery for 'list_items', which has the `item_type_1`/2/3
> connected to it as a joinedload
>
> the closest I can get is this:
>
> query = s.query(List)\
> .options(
> subqueryload('list_item')\
> .joinedload('item_type_1')
> )
>
> at that point, the unbound subquery can only join things from `item_type_1`.
> i can't figure out how to join other relationships of `list_item`
>
> is it possible to joinedload mutliple trees/paths onto a subqueryload ?
in the old way, you’d say:
subqueryload(‘list_item’),
joinedload('list_item.item_type_1'),
joinedload('list_item.item_type_2'),
joinedload('list_item.item_type_3'),
that will still work.
new way you can also say:
subqueryload('list_item').joinedload('item_type_1'),
defaultload('list_item').joinedload(item_type_2'),
defaultload('list_item').joinedload('item_type_3'),
or variants thereof.
>
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