fastapi自定义响应类作为默认响应类

如何解决fastapi自定义响应类作为默认响应类

我正在尝试将自定义响应类用作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",**kwargs,)

当我明确使用响应类时，这很好用。

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

但是，当我尝试将此参数作为参数传递给FastAPI构造函数时，似乎只在从请求处理程序返回数据时才使用它。

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，似乎它仍使用正常的默认响应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,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,send)
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/starlette/middleware/errors.py",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,_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,sender)
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/starlette/routing.py",line 566,in __call__
    await route.handle(scope,send)
  File "/home/blue/podman/test/.venv/lib/python3.6/site-packages/starlette/routing.py",line 227,in handle
    await self.app(scope,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',)]

解决方法

因此，事实证明，默认响应类以及路由上的响应类仅适用于开放API文档。默认情况下，文档将记录每个端点，就像它们将返回json一样。

因此，在下面的示例代码中，每个响应都将被标记为内容类型text / html。 在第二次溃败中，这被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"}

从这种意义上讲，我可能应该显式使用我的类，并将默认响应类保留为JSON，以便将它们记录为JSON响应。