title: The "correct" way to define an exception in Python without PyLint complaining
body: <p>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.</p>
  
  <p>First, the simplest way:</p>
  
  <pre><code>class MyException(Exception):
      pass
  </code></pre>
  
  <p>This works, but prints out a warning at runtime: <a href="https://stackoverflow.com/questions/1272138/baseexception-message-deprecated-in-python-2-6">DeprecationWarning: BaseException.message has been deprecated as of Python 2.6</a> OK, so that's not the way. I then tried:</p>
  
  <pre><code>class MyException(Exception):
      def __init__(self, message):
          self.message = message
  </code></pre>
  
  <p>This also works, but PyLint reports a warning: <code>W0231: MyException.__init__: __init__ method from base class 'Exception' is not called</code>. So I tried calling it:</p>
  
  <pre><code>class MyException(Exception):
      def __init__(self, message):
          super(Exception, self).__init__(message)
          self.message = message
  </code></pre>
  
  <p>This works, too! But now PyLint reports an error: <code>E1003: MyException.__init__: Bad first argument 'Exception' given to super class</code></p>
  
  <p>How the hell do I do such a simple thing without any warnings?</p>