title: How to get the current running module path/name
body: <p>I've searched and this seems to be a simple question without a simple answer.</p>
  <p>I have the file <code>a/b/c.py</code> which would be called with <code>python -m a.b.c</code>. I would like to obtain the value <code>a.b.c</code> in the module level.</p>
  <pre class="lang-py prettyprint-override"><code>USAGE = u'''\
  Usage:
      python -m %s -h
  ''' % (what_do_i_put_here,)
  </code></pre>
  <p>So when I receive the <code>-h</code> option, I display the <code>USAGE</code> without the need to actually write down the actual value in each and every script.</p>
  <p>Do I really need to go through <code>inspect</code> to get the desired value?</p>
  <p>Thanks.</p>
  <p><strong>EDIT:</strong> As said, there are answers (I've searched), but not simple answers. Either use <code>inspect</code>, use of <code>traceback</code>, or manipulate <code>__file__</code> and <code>__package__</code> and do some substring to get the answer. But nothing as simple as if I had a class in the module, I could just use <code>myClass.__module__</code> and I would get the answer I want. The use of <code>__name__</code> is (unfortunately) useless as it's always <code>&quot;__main__&quot;</code>.</p>
  <p>Also, this is in python 2.6 and I cannot use any other versions.</p>