python empty argument

Question

Hello,

how do I print help info if no arguments are passed to python script?

#!/usr/bin/env python

import sys

for arg in sys.argv:
    if arg == "do":
        do this
    if arg == ""
        print "usage is bla bla bla"

what I'm missing is if arg == "" line that I don't know how to express :(

Answer 1

if len(sys.argv)<2:

The name of the program is always in sys.argv[0]

Answer 2

if len(sys.argv) == 1:
    # Print usage...

The first element of sys.argv is always either the name of the script itself, or an empty string. If sys.argv has only one element, then there must have been no arguments.

http://docs.python.org/library/sys.html#sys.argv

Answer 3

#!/usr/bin/env python

import sys
args = sys.argv[1:]

if args:
    for arg in args:
        if arg == "do":
            # do this
else:
    print "usage is bla bla bla"

Answer 4

Based on Noctis Skytower's answer

import sys
args = sys.argv[1:]

for arg in args:
    if arg == "do":
        # do this

if not args:
    print "usage is bla bla bla"

Answer 5

As others have said, you can check if any args were passed in by doing:

#!/usr/bin/env python

import sys
args = sys.argv[1:]

if args:
    for arg in args:
        if arg == "do":
            # do this
else:
    print "usage is bla bla bla"

However, there is a Python module called OptParse that was developed explicitly for parsing command line arguments when running a script. I would suggest looking into this, as it's a bit more "standards compliant" (As in, it's the expected and accepted method of command line parsing within the Python community).

Answer 6

I recommend you use the lib optparse [1], is more elegant :D

[1] More powerful command line option parser < http://docs.python.org/library/optparse.html >