無法將響應返回給瀏覽器 flask

Question

我有一個 web 服務，它應該保持監聽實時聊天的新消息，問題是我無法返回它的響應或簡單的文本，因為這部分代碼：

    # Print every message the current user would receive
     # This is a blocking call that will run forever
    (client.call_on_each_message(debug))

永遠保持傾聽，所以我實現了一個實際的 function 來返回響應，但我的代碼以打印測試結束，沒有通過返回消息，這行代碼：

    return jsonify((str(msg))), 200

我該怎么做才能將響應返回給瀏覽器

    @app.route('/listenrealtime', methods=['GET'])
    def listenrealtime():
         def debug(msg): 
             msg: lambda msg: (str(msg))
             print('test')
             print(msg)
             return jsonify((str(msg))), 200
       
         # Pass the path to your zuliprc file here.
         client = zulip.Client(config_file="~/zuliprc")
     
         # Print every message the current user would receive
         # This is a blocking call that will run forever
         (client.call_on_each_message(debug))
     
         # Print every event relevant to the user
         # This is a blocking call that will run forever
         return "Some text"

Answer 1

它由於錯位的括號而無法返回：

return jsonify(((str(msg)), 200)