Python：改造flask接口服务提供统一格式的返回数据

TypeError: The view function did not return a valid response. 
The return type must be a string, dict, tuple, Response instance, or WSGI callable, 
but it was a JsonResponse.

# -*- coding: utf-8 -*-
from flask import Flask, jsonify
from json_response import JsonResponse
class JsonFlask(Flask):
    def make_response(self, rv):
        """视图函数可以直接返回: list、dict、None"""
        if rv is None or isinstance(rv, (list, dict)):
            rv = JsonResponse.success(rv)
        if isinstance(rv, JsonResponse):
            rv = jsonify(rv.to_dict())
        return super().make_response(rv)

# -*- coding: utf-8 -*-
class JsonResponse(object):
    """
    统一的json返回格式
    """
    def __init__(self, data, code, msg):
        self.data = data
        self.code = code
        self.msg = msg
    @classmethod
    def success(cls, data=None, code=0, msg='success'):
        return cls(data, code, msg)
    @classmethod
    def error(cls, data=None, code=-1, msg='error'):
        return cls(data, code, msg)
    def to_dict(self):
        return {
            "code": self.code,
            "msg": self.msg,
            "data": self.data
        }

# -*- coding: utf-8 -*-
from flask import Flask
from flask_cors import CORS
from json_flask import JsonFlask
from json_response import JsonResponse
# app = Flask(__name__)
app = JsonFlask(__name__)
CORS(app, supports_credentials=True)
@app.route("/")
def index():
    return ['Tom', 'Jack']
@app.errorhandler(Exception)
def error_handler(e):
    """
    全局异常捕获，也相当于一个视图函数
    """
    return JsonResponse.error(msg=str(e))
if __name__ == '__main__':
    app.run(port=5001)

{
    "code": 0,
    "data": [
        "Tom",
        "Jack"
    ],
    "msg": "success"
}