Add encoder argument to JSONResponse

[alex-oleshkevich]

This small change allows users to serialize a custom types with a little effort compared to subclassing.

import datetime
import json

from starlette.responses import JSONResponse


class MyJSONEncoder(json.JSONEncoder):
    def default(self, value):
        if isinstance(value, datetime.datetime):
            return value.isoformat()
        ...

async def app(scope, receive, send):
    assert scope['type'] == 'http'
    response = JSONResponse({'now': datetime.datetime.now()}, encoder=MyJSONEncoder)
    await response(scope, receive, send)

Similar to Django - https://docs.djangoproject.com/en/4.2/ref/request-response/#jsonresponse-objects

[Kludex]

We already had this discussion before... Need to find the reference.

But... I'm not keen on this change. It's too easy for the user to implement its custom Response class.

[alex-oleshkevich]

I found it cleaner to share the encoder (when used not only for the API JSON responses) than building a custom response class. Moreover, inheriting the base class requires you to completely override JSON serialization, which may be unwanted in cases when base JSONResponse class does more than just json.dumps (not a case for now).

[Kludex]

• Using Custom JSON Encoders and Decoders? #1517
• Could JSONResponse provide JsonEncoder class option on JSONResponse.__init__ () ? #1281

[Kludex]

This PR goes against what is written in the "Custom JSON serialization" section in the documentation (the above section).

[alex-oleshkevich]

ok, closing it