Find module name of the originating exception in Python

Question

Example:

>>> try:
...    myapp.foo.doSomething()
... except Exception, e:
...    print 'Thrown from:', modname(e)

Thrown from: myapp.util.url

In the above example, the exception was actually thrown at myapp/util/url.py module. Is there a way to get the __name__ of that module?

My intention is to use this in logging.getLogger function.

Answer 1

This should work:

import inspect

try:
    some_bad_code()
except Exception, e:
    frm = inspect.trace()[-1]
    mod = inspect.getmodule(frm[0])
    print 'Thrown from', mod.__name__

EDIT: Stephan202 mentions a corner case. In this case, I think we could default to the file name.

import inspect

try:
    import bad_module
except Exception, e:
    frm = inspect.trace()[-1]
    mod = inspect.getmodule(frm[0])
    modname = mod.__name__ if mod else frm[1]
    print 'Thrown from', modname

The problem is that if the module doesn't get loaded (because an exception was thrown while reading the code in that file), then the inspect.getmodule call returns None. So, we just use the name of the file referenced by the offending frame. (Thanks for pointing this out, Stephan202!)

Answer 2

You can use the traceback module, along with sys.exc_info(), to get the traceback programmatically:

try:
    myapp.foo.doSomething()
except Exception, e:
    exc_type, exc_value, exc_tb = sys.exc_info()
    filename, line_num, func_name, text = traceback.extract_tb(exc_tb)[-1]
    print 'Thrown from: %s' % filename