python empty argument

user_264532 0
Viewed 62978

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 :(

parameter-passing arguments python
8 Answers
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

user_138270 
if len(sys.argv)<2:

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

user_1993970

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 are Python modules called argparse or OptParse (now deprecated) that were 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).

user_1176030 
#!/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"
user_2163560

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"
user_1747280
user_1176820
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