我们使用geventwebsocket和javascript时，flask app.py无法正常工作

如何解决我们使用geventwebsocket和javascript时，flask app.py无法正常工作

我正在从摄像头捕获帧。为此，我使用了两个2个文件。 streamer.html和app.py。我写了这个streamer.html捕获帧。在app.py中，我编写了通过websocket捕获框架的代码。但是它失败了。我正在使用gevent-websocket。当我们使用geventwebsocket app.py无法在Flask中运行时。

from geventwebsocket.handler import WebSocketHandler
from gevent.pywsgi import WsgiServer
from flask import Flask,request,render_template

app = Flask(__name__)

@app.route('/api')
def api():
    if request.environ.get('wsgi.websocket'):
        ws = request.environ['wsgi.websocket']
        while True:
            message = ws.receive()
            print(message)
            ws.send(message)
    else:
        print('Hello')

    return

if __name__ == '__main__':
    http_server = WsgiServer(("",5000),api,handler_class=WebSocketHandler)
    http_server.serve_forever()

而Javascript代码是

const ws = new WebSocket（“ ws：// localhost：5000 / api”）

  // Stream video frames are byte64 image to server
  ws.onopen = () => {
    console.log(`Connected to ${WS_URL}`);
    setInterval(() => {
      const streamData = {
        frame: getFrame(),};
      ws.send(JSON.stringify(streamData));
    },1000 / FPS);
  };

  // When data is received
  ws.onmessage = function (event) {
    // Repaint the Feedback div with updated data from server. Ideally it will have processed data from server.
    const FeedbackDiv = document.getElementById("showFeedbackToUser");
    const receivedMessage = JSON.parse(event.data);
    FeedbackDiv.innerHTML =
      "Your Sr No: " +
      receivedMessage.clientSrNo +
      " Your dynamic name: " +
      receivedMessage.name;
  };