fastapi 自定义响应 class 作为默认响应 class

Question

我正在尝试将自定义响应 class 用作 default response。

from fastapi.responses import Response
from bson.json_util import dumps

class MongoResponse(Response):
    def __init__(self, content, *args, **kwargs):
        super().__init__(
            content=dumps(content),
            media_type="application/json",
            *args,
            **kwargs,
        )

当我明确使用响应 class 时，这工作正常。

@app.get("/")
async def getDoc():
    foo = client.get_database('foo')
    result = await foo.bar.find_one({'author': 'fool'})
    return MongoResponse(result)

但是，当我尝试将它作为参数传递给 FastAPI 构造函数时，它似乎在仅从请求处理程序返回数据时没有被使用。

app = FastAPI(default_response_class=MongoResponse)

@app.get("/")
async def getDoc():
    foo = client.get_database('foo')
    result = await foo.bar.find_one({'author': 'fool'})
    return result

当我查看下面的堆栈跟踪时，它似乎仍在使用正常的默认响应，即 json response。

ERROR:    Exception in ASGI application
Traceback (most recent call last):
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/uvicorn/protocols/http/httptools_impl.py", line 390, in run_asgi
    result = await app(self.scope, self.receive, self.send)
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/uvicorn/middleware/proxy_headers.py", line 45, in __call__
    return await self.app(scope, receive, send)
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/fastapi/applications.py", line 181, in __call__
    await super().__call__(scope, receive, send)  # pragma: no cover
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/starlette/applications.py", line 111, in __call__
    await self.middleware_stack(scope, receive, send)
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/starlette/middleware/errors.py", line 181, in __call__
    raise exc from None
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/starlette/middleware/errors.py", line 159, in __call__
    await self.app(scope, receive, _send)
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/starlette/exceptions.py", line 82, in __call__
    raise exc from None
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/starlette/exceptions.py", line 71, in __call__
    await self.app(scope, receive, sender)
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/starlette/routing.py", line 566, in __call__
    await route.handle(scope, receive, send)
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/starlette/routing.py", line 227, in handle
    await self.app(scope, receive, send)
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/starlette/routing.py", line 41, in app
    response = await func(request)
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/fastapi/routing.py", line 199, in app
    is_coroutine=is_coroutine,
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/fastapi/routing.py", line 122, in serialize_response
    return jsonable_encoder(response_content)
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/fastapi/encoders.py", line 94, in jsonable_encoder
    sqlalchemy_safe=sqlalchemy_safe,
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/fastapi/encoders.py", line 139, in jsonable_encoder
    raise ValueError(errors)
ValueError: [TypeError("'ObjectId' object is not iterable",), TypeError('vars() argument must have __dict__ attribute',)]

Answer 1

所以事实证明，默认响应 class 以及路由上的响应 class 仅适用于打开的 API 文档。默认情况下，文档将记录每个端点，就好像它们 return json.

因此，对于下面的示例代码，每个响应都将被标记为内容类型 text/html。在第二次溃败时，它被 application/json

覆盖

app = FastAPI(default_response_class=HTMLResponse)

@app.get("/")
async def getDoc():
    foo = client.get_database('foo')
    result = await foo.bar.find_one({'author': 'Mike'})
    return MongoResponse(result)


@app.get("/other", response_class=JSONResponse)
async def json():
    return {"json": "true"}

从这个意义上说，我可能应该明确使用我的 class 并将默认响应 class 保留为 JSON，以便将它们记录为 JSON 响应。

fastapi 自定义响应 class 作为默认响应 class

fastapi custom response class as default response class

python

python-3.x

fastapi