How do I get the method that my exception was raised in?

Question

I have various error checking methods and they are mainly just value or type checking and I want to give the user a chance to fix it so they don't lose a bunch of information regarding what the program is doing.

At this point, I just have this :

def foo(Option1, Option2): 
    if Option1 >= 0 and Option <= 100 :
        continue
    else:
        e = ('Hey this thing doesn\'t work')
        raise ValueError(e)

and then later in the program that is calling it, I have

except ValueError as e:
    print(e)

I want to pass what method was the problem so that I can give the user a chance to try again, like with a prompt or something after right around where the print(e) statement is. Any ideas?

Edit:

Basically I would like my except code to look something like this

except ValueError as e:
    # print the error
    # get what method the error was raised in
    # method = the_method_from_above
    # prompt user for new value
    # send command to the method using the new value

Answer 1

You can use the traceback module to provide stack trace information about exceptions.

import traceback
...
try:
  pass
except ValueError as e:
  print("Error {0}".format(e))
  traceback.print_exc()

Answer 2

You can do this with some introspection, but you shouldn't.

The following code will let you call the function in which the exception was raised, but there's almost certainly a better way to do whatever it is you're trying to achieve...

import sys

def foo(x):
    print('foo(%r)' % x)
    if not (0 <= x <= 100):
        raise ValueError

def main():
    try:
        foo(-1)
    except ValueError:
        tb = sys.exc_info()[2]
    while tb.tb_next is not None:
        tb = tb.tb_next
    funcname = tb.tb_frame.f_code.co_name
    func = globals()[funcname]
    func(50)

if __name__ == '__main__':
    main()

...which prints out...

foo(-1)
foo(50)