How Do I Get the Module Name of an Object's Class Definition Rather Than the Module Name of the Object's Instantiation?

Question

In python 2.5, I have the following code in a module called modtest.py:

def print_method_module(method):     def printer(self):         print self.__module__         return method(self)     return printer  class ModTest():      @print_method_module     def testmethod(self):         pass  if __name__ == "__main__":     ModTest().testmethod() 

However, when I run this, it prints out:

__main__ 

If I create a second file called modtest2.py and run it:

import modtest  if __name__ == "__main__":     modtest.ModTest().testmethod() 

This prints out:

modtest 

How can I change the decorator to always print out modtest, the name of the module in which the class is defined?

Answer 1

When you execute a python source file directly, the module name of that file is __main__, even if it is known by another name when you execute some other file and import it.

You probably want to do like you did in modtest2, and import the module containing the class definition instead of executing that file directly. However, you can get the filename of the main module like so, for your diagnostic purposes:

def print_method_module(method):     def printer(self):         name = self.__module__         if name == '__main__':             filename = sys.modules[self.__module__].__file__             name = os.path.splitext(os.path.basename(filename))[0]         print name         return method(self)     return printer