Find Module Name Of The Originating Exception In Python

Example: >>> try: ... myapp.foo.doSomething() ... except Exception, e: ... print 'Thrown from:', modname(e) Thrown from: myapp.util.url In the above example, the e

Solution 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!)

Solution 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

Solution 3:

This should do the trick:

import inspect
def modname():
    t=inspect.trace()
    if t:
        return t[-1][1]

Solution 4:

Python's logging package already supports this - check the documentation. You just have to specify %(module)s in the format string. However, this gives you the module where the exception was caught - not necessarily the same as the one where it was raised. The traceback, of course, gives you the precise location where the exception was raised.

Solution 5:

I have a story about how CrashKit computes class names and package names from Python stack traces on the company blog: “Python stack trace saga”. Working code included.