为什么我的 Flask 错误处理程序没有被调用？

Question

我正在尝试使用“Implementing API Exceptions”页面上的指南在 Flask 1.0.2 和 Flask-RESTful 0.3.7 中创建自定义错误处理程序。 （Flask-RESTful 有自己的creating custom error messages 方式，但由于它似乎没有办法在异常时接受自定义错误消息，所以我尝试使用 vanilla Flask 方法。）

from flask import Flask, jsonify
from flask_restful import Resource, Api

app = Flask(__name__)
api = Api(app)

#########################################

class MyGenericException(Exception):
    status_code = 500
    def __init__(self, message, status_code=None, payload=None):
        Exception.__init__(self)
        self.message = message
        if status_code is not None:
            self.status_code = status_code
        self.payload = payload
    def to_dict(self):
        rv = dict(self.payload or ())
        rv['message'] = self.message
        return rv

@app.errorhandler(MyGenericException)
def handle_generic_error(error):
    response = jsonify(error.to_dict())
    response.status_code = error.status_code
    return response

#########################################

class TestMe(Resource):
    def get(self):
        raise MyGenericException('A generic error', status_code=501)
api.add_resource(TestMe, '/testme', endpoint='TestMe')

#########################################

if __name__ == '__main__':
    app.run(debug=False)

调用http://127.0.0.1:5000/testme 只会返回一个通用的“500 内部服务器错误”消息，但不会返回带有我的自定义错误文本的 501 错误。似乎MyGenericException 被正确提升，但 Flask 似乎忽略了它。

[2019-05-08 17:09:18,409] ERROR in app: Exception on /testme [GET]
Traceback (most recent call last):
  File "C:\Users\testuser\Envs\testenv\lib\site-packages\flask\app.py", line 1813, in full_dispatch_request
    rv = self.dispatch_request()
  File "C:\Users\testuser\Envs\testenv\lib\site-packages\flask\app.py", line 1799, in dispatch_request
    return self.view_functions[rule.endpoint](**req.view_args)
  File "C:\Users\testuser\Envs\testenv\lib\site-packages\flask_restful\__init__.py", line 458, in wrapper
    resp = resource(*args, **kwargs)
  File "C:\Users\testuser\Envs\testenv\lib\site-packages\flask\views.py", line 88, in view
    return self.dispatch_request(*args, **kwargs)
  File "C:\Users\testuser\Envs\testenv\lib\site-packages\flask_restful\__init__.py", line 573, in dispatch_request
    resp = meth(*args, **kwargs)
  File "C:/Users/testuser/Documents/PyCharm Projects/TestApp/testapp.py", line 32, in get
    raise MyGenericException('A generic error', status_code=505)
MyGenericException

@app.errorhandler 装饰器似乎已为自定义 MyGenericException 异常正确设置。为什么 Flask 不处理它？

感谢任何可以提供帮助的人。

Answer 1

参考this question 和docs，我从两者中获得的关键信息是，如果您的路线是 Flask-RESTful 路线，您的路线将由handle_error() 处理，并且防止或自定义这种情况的唯一方法是实现您自己的 API 类，并覆盖 handle_error()。

Answer 2

跟进@dylanj.nz 的回答this vanilla-Flask error handling method 和this example of overriding Flask-RESTful's API，这是我确定的方法。它允许 Flask-RESTful 处理 HTTPException 类型的异常，但将其他所有内容传递给默认 (Flask) 处理程序，在该处理程序中可以指定自定义错误消息（键/值条目的整个 JSON 对象，如果需要）异常发生的时间。

from flask_restful import Resource, Api as _Api, HTTPException
app = Flask(__name__)

# This new Exception will accept a message, a status code, and a
# payload of other values to be displayed as a JSON object
class FlaskGenericException(Exception):
    status_code = 500   # default unless overridden
    def __init__(self, message, status_code=None, payload=None):
        Exception.__init__(self)
        self.message = message
        if status_code is not None:
            self.status_code = status_code
        self.payload = payload
    def to_dict(self):
        rv = dict(self.payload or ())
        rv['message'] = self.message
        return rv

@app.errorhandler(FlaskGenericException)
def handle_flask_generic_error(error):
    response = jsonify(error.to_dict())
    response.status_code = error.status_code
    return response

# This overridden Flask-RESTful API class will keep Flask-RESTful
# from handling errors other than HTTPException ones.
class Api(_Api):
    def error_router(self, original_handler, e):
        # Override original error_router to only handle HTTPExceptions.
        if self._has_fr_route() and isinstance(e, HTTPException):
            try:
                # Use Flask-RESTful's error handling method
                return self.handle_error(e) 
            except Exception:
                # Fall through to original handler (i.e. Flask)
                pass
        return original_handler(e)

api = Api(app)

class TestMe(Resource):
    def get(self):
        try:
            ldapc = ldap.connection
        except:
            # message = first parameter.  Other parameters come in as "payload"
            raise FlaskGenericException('A generic error', status_code=505, payload={'user': 'John Doe', 'company': 'Foobar Corp.'})

api.add_resource(TestMe, '/testme', endpoint='TestMe')