快速的api，在Web套接字中发送json响应

如何解决快速的api，在Web套接字中发送json响应

我有一个使用快速api构建的小型WebSocket。 我正在尝试从客户端接收消息，对其进行处理，然后返回JSON响应。

@app.websocket("/ws")
async def websocket_endpoint(websocket: WebSocket):
    await websocket.accept()
    while True:
        story = await websocket.receive_text()
        data = sentiment_analysis(news=story)
# Exmaple out put of sentiment_analysis function: {'compound_score': 0.5856095238095237,'correlation_score': 0.8736042200587304}

        await websocket.send_json(json.dumps(data))

来自客户端的消息将按应有的方式进行接收和处理，但是我似乎无法将响应字典加载到JSON并将其发送回去。

我遇到错误：

  ERROR:    Exception in ASGI application
Traceback (most recent call last):
  File "/usr/lib/python3/dist-packages/uvicorn/protocols/websockets/websockets_impl.py",line 153,in run_asgi
    result = await self.app(self.scope,self.asgi_receive,self.asgi_send)
  File "/usr/lib/python3/dist-packages/uvicorn/middleware/proxy_headers.py",line 45,in __call__
    return await self.app(scope,receive,send)
  File "/usr/local/lib/python3.8/dist-packages/fastapi/applications.py",line 179,in __call__
    await super().__call__(scope,send)
  File "/usr/local/lib/python3.8/dist-packages/starlette/applications.py",line 111,in __call__
    await self.middleware_stack(scope,send)
  File "/usr/local/lib/python3.8/dist-packages/starlette/middleware/errors.py",line 146,in __call__
    await self.app(scope,send)
  File "/usr/local/lib/python3.8/dist-packages/starlette/exceptions.py",line 58,send)
  File "/usr/local/lib/python3.8/dist-packages/starlette/routing.py",line 566,in __call__
    await route.handle(scope,line 283,in handle
    await self.app(scope,line 57,in app
    await func(session)
  File "/usr/local/lib/python3.8/dist-packages/fastapi/routing.py",line 228,in app
    await dependant.call(**values)
  File "./app.py",line 20,in websocket_endpoint
    stock = await websocket.receive_text()
  File "/usr/local/lib/python3.8/dist-packages/starlette/websockets.py",line 85,in receive_text
    self._raise_on_disconnect(message)
  File "/usr/local/lib/python3.8/dist-packages/starlette/websockets.py",line 80,in _raise_on_disconnect
    raise WebSocketDisconnect(message["code"])
starlette.websockets.WebSocketDisconnect: 1011

解决方法

如果您使用的是 websocket.send_json()，那么您不需要执行 json.dumps()，因为这只是将其解析为字符串文本。

我还发现 @app.websocket() 对我不起作用，我需要使用 websocket_route() 装饰器。 FastAPI 一直抛出 WebSocketDisconnect，直到我使用 websocket_route() 装饰器。

注意：我使用的是带有 APIRouter() 而不是 FastAPI() 的装饰器，所以请稍加保留，但我相信最新版本的 FastAPI 支持 websocket_route()。

下面的代码应该适用于这 2 个修改

---

@app.websocket_route("/ws")
async def websocket_endpoint(websocket: WebSocket):
    await websocket.accept()
    while True:
        story = await websocket.receive_text()
        data = sentiment_analysis(news=story)
        await websocket.send_json(data)