How to get the caller script name

Question

I\'m using Python 2.7.6 and I have two scripts:

outer.py

import sys
import os

print \"Outer file launching...\"
os.system(\'inner.py\')

Answer 1

If applicable to your situation you could also simply pass an argument that lets inner.py differentiate:

import sys
import os

print "Outer file launching..."
os.system('inner.py launcher')

innter.py

import sys
import os

try:
    if sys.argv[0] == 'launcher':
        print 'outer.py called us'
except:
    pass