Is it possible to instantiate the mapped strawberry classes? about strawberry-sqlalchemy HOT 6 OPEN

Possibly you can explicitly invoke Strawberry's field resolution (it's been a while since I've taken a look, but hopefully it's in the docs, otherwise you can dig into the source), but I'd personally just typing.cast() it.

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My problem isn't with the type checker but with the actual runtime (don't want to have to call StrawberryType.my_field(obj) everywhere, would much rather do obj.my_field()), so typing.cast won't do it.

I can do an absolutely hideous
result.__class__ = StrawberryType

but I find that a little scary

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Also, it seems like the autogenerated relationship resolver returns the model class instead of the strawberry class. Would love if it would return the strawberry class instead

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@TimDumol

I got working locally a version of strawberry-sqlalchemy-mapper that generates a from_row classmethod and returns strawberry types instead of sqlalchemy types from each resolver. The resolvers can also handle a model being passed in as self for backwards compatibility, but I think the documentation should encourage returning strawberry types (else the static methods get confusing fast).

If I sent you a PR, do you think you could help me with some of the finer details? Namely, I don't know what a mapped interface is (as opposed to a type) and wouldn't be surprised if this fails for that case. I wouldn't be surprised if there are some other edge cases I missed.

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Hey @mattalbr -- sorry, been fairly busy recently. I'm not entirely certain what you're trying to achieve, but I believe a PR would help communicate that more effectively, so go for it.

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Sure thing! Sent the pull request, with a bunch of my thoughts in the description. Please take a look when you get a chance, and we can discuss/iterate over there.

#16

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