Open Bingdom opened 1 year ago
That's a use case we haven't considered. You can possibly try something like:
from dataclasses import make_dataclass
@_strawberryMapper.type(schema.User)
class User:
pass
_strawberryMapper.finalize()
UserInput = strawberry.input(make_dataclass('UserInput', [], bases=(_strawberryMapper.mapped_types['User'],))
Haven't actually tried this code, but maybe it will give you a good starting point.
Hi,
Thanks for the help.
Strangely, I'm getting the same problem.
What would be the main difference writing like that vs naturally? Except I suppose the additional methods dataclass provides.
I suspect there might be a property that doesn't get correctly overridden that strawberry uses to identify the type?
Hi, @Bingdom, you can use @strawberry.input
with @strawberry_sqlalchemy_mapper.type
.
Like this:
@strawberry.input
@strawberry_sqlalchemy_mapper.type(User)
class UserInput:
__exclude__ = ["id", "password_hash"]
This help me with that limitation, I hope it can help you too!
Hello,
I was wondering if there's a way to map to an input type?
In Strawberry, you could just do:
But doing that with this library doesn't make strawberry recognize it as an input
Error:
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