Python unittest.TestCase execution order

Don't make them independent tests - if you want a monolithic test, write a monolithic test.

class Monolithic(TestCase):
  def step1(self):
      ...

  def step2(self):
      ...

  def _steps(self):
    for name in dir(self): # dir() result is implicitly sorted
      if name.startswith("step"):
        yield name, getattr(self, name) 

  def test_steps(self):
    for name, step in self._steps():
      try:
        step()
      except Exception as e:
        self.fail("{} failed ({}: {})".format(step, type(e), e))

If the test later starts failing and you want information on all failing steps instead of halting the test case at the first failed step, you can use the subtests feature: https://docs.python.org/3/library/unittest.html#distinguishing-test-iterations-using-subtests

(The subtest feature is available via unittest2 for versions prior to Python 3.4: https://pypi.python.org/pypi/unittest2 )

From unittest — Unit testing framework, section Organizing test code:

Note: The order in which the various tests will be run is determined by sorting the test method names with respect to the built-in ordering for strings.

So just make sure test_setup's name has the smallest string value.

Note that you should not rely on this behavior — different test functions are supposed to be independent of the order of execution. See ngcohlan's answer above for a solution if you explicitly need an order.

It's a good practice to always write a monolithic test for such expectations. However, if you are a goofy dude like me, then you could simply write ugly looking methods in alphabetical order so that they are sorted from a to b as mentioned in the Python documentation - unittest — Unit testing framework

Note that the order in which the various test cases will be run is determined by sorting the test function names with respect to the built-in ordering for strings

Example

  def test_a_first():
  print "1"
  def test_b_next():
  print "2"
  def test_c_last():
  print "3"