The "correct" way to define an exception in Python without PyLint complaining

Question

I'm trying to define my own (very simple) exception class in Python 2.6, but no matter how I do it I get some warning.

First, the simplest way:

class MyException(Exception):     pass 

This works, but prints out a warning at runtime: DeprecationWarning: BaseException.message has been deprecated as of Python 2.6 OK, so that's not the way. I then tried:

class MyException(Exception):     def __init__(self, message):         self.message = message 

This also works, but PyLint reports a warning: W0231: MyException.__init__: __init__ method from base class 'Exception' is not called. So I tried calling it:

class MyException(Exception):     def __init__(self, message):         super(Exception, self).__init__(message)         self.message = message 

This works, too! But now PyLint reports an error: E1003: MyException.__init__: Bad first argument 'Exception' given to super class

How the hell do I do such a simple thing without any warnings?

Answer 1

When you call super, you need the subclass/derived class as the first argument, not the main/base class.

From the Python online documentation:

class C(B):     def method(self, arg):         super(C, self).method(arg) 

So your exception would be defined as follows:

class MyException(Exception):     def __init__(self, message):         super(MyException, self).__init__(message)         self.message = message