Question

I am developing a Python module with several source files, each with its own test class derived from unittest right in the source. Consider the directory structure:

dirFoo\     test.py     dirBar\         __init__.py         Foo.py         Bar.py 

To test either Foo.py or Bar.py, I would add this at the end of the Foo.py and Bar.py source files:

if __name__ == '__main__':     unittest.main() 

And run Python on either source, i.e.

$ python Foo.py ........... ---------------------------------------------------------------------- Ran 11 tests in 2.314s  OK 

Ideally, I would have ‘test.py’ automagically search dirBar for any unittest derived classes and make one call to ‘unittest.main()’. What’s the best way to do this in practice?

I tried using Python to call execfile for every *.py file in dirBar, which runs once for the first .py file found & exits the calling test.py, plus then I have to duplicate my code by adding unittest.main() in every source file–which violates DRY principles.

Answer 1

I knew there was an obvious solution:

dirFoo\     __init__.py     test.py     dirBar\         __init__.py         Foo.py         Bar.py 

Contents of dirFoo/test.py

from dirBar import * import unittest  if __name__ == "__main__":      unittest.main() 

Run the tests:

$ python test.py ........... ---------------------------------------------------------------------- Ran 11 tests in 2.305s  OK