How to pass command line options using python's command line shell? [duplicate]

Question

This is my program(test.py):

#!/usr/bin/python
import sys, getopt
def main(argv):
   inputfile = ''
   outputfile = ''
   try:
      opts, args = getopt.getopt(argv,"hi:o:",["ifile=","ofile="])
   except getopt.GetoptError:
      print 'test.py -i <inputfile> -o <outputfile>'
      sys.exit(2)
   for opt, arg in opts:
      if opt == '-h':
         print 'test.py -i <inputfile> -o <outputfile>'
         sys.exit()
      elif opt in ("-i", "--ifile"):
         inputfile = arg
      elif opt in ("-o", "--ofile"):
         outputfile = arg
   print 'Input file is "', inputfile
   print 'Output file is "', outputfile

if __name__ == "__main__":
     main(sys.argv[1:])

Using msdos command line I could pass the -h option(defined in test.py) like this:

python test.py -h

The msdos command line would then ouput this:

 test.py -i <inputfile> -o <outputfile>

But how would i pass the -h option in python interactive mode as was done using the msdos command line?

elyase · Accepted Answer · 2013-08-18 00:22:10Z

1

May be you could try to hack something with a custom sys.argv but this would be way too hacky, use instead:

>>> from subprocess import call
>>> call(['./test.py', option1, option2, ...])

edited Aug 18, 2013 at 0:22

answered Aug 18, 2013 at 0:10

elyase

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rodrigo · Accepted Answer · 2013-08-18 01:03:05Z

0

The whole idea of the if __name__ == "__main__": line is that this file can be used both as a program and as a module.

So do this:

>>> import test
>>> test.main(['-h'])

If your module wouldn't have the __name__ check you could do just assigning to sys.argv:

>>> import sys
>>> sys.argv = ['-h']
>>> import test

But naturally that would only work the first time the module is loaded. For the next runs you will need to run:

>>> reload(test)

Note: In Python2 reload is a built-in, but in Python3 it is in module imp.

edited Aug 18, 2013 at 1:03

answered Aug 18, 2013 at 0:56

rodrigo

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