Question

Python has a nice execfile function inside the interpreter where you can run a program, keep all the variables in scope, and then inspect them at your leisure. However, as far as I know you can’t run execfile on a program that takes arguments from the command line; if you try to include the arguments, Python throws an IOError and complains that the file (with spaces and arguments) can’t be found.

Is there any way to run a Python script that takes command line arguments, and keep all of the variables in scope after the program executes? Like an execfile that takes flags?

Thanks,
Kevin

Answer 1

You could modify sys.argv directly. The file:

# foo.py
import sys

print sys.argv

The other file:

import sys
import shlex   # thanks Matt

old_argv = sys.argv
sys.argv = shlex.split('foo.py is a happy camper')

execfile('foo.py')

Output:

$ python foo.py is a happy camper
['foo.py', 'is', 'a', 'happy', 'camper']
$ python bar.py 
['foo.py', 'is', 'a', 'happy', 'camper']

But I must say, quoting Ignacio Vazquez-Abrams:

This sounds all sorts of messed up.

I’m assuming you have your reasons though.