无法将响应返回给浏览器 flask

Question

我有一个 web 服务，它应该保持监听实时聊天的新消息，问题是我无法返回它的响应或简单的文本，因为这部分代码：

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

永远保持倾听，所以我实现了一个实际的 function 来返回响应，但我的代码以打印测试结束，没有通过返回消息，这行代码：

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

我该怎么做才能将响应返回给浏览器

    @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

它由于错位的括号而无法返回：

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