python command line arguments in main, skip script name

Question

This is my script

def main(argv):
    if len(sys.argv)>1:
        for x in sys.argv:
            build(x)

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

so from the command line I write python myscript.py commandlineargument

I want it to skip myscript.py and simply run commandlineargument through commandlineargument(n)

so I understand that my for loop doesn't account for this, but how do I make it do that?

Eldreeda

If the main function takes an argv parameter, it should probably use that parameter, rather than ignoring it and using sys.argv instead… – Pugilism

Also, you don't need the if check at all. If there are no arguments, the loop will successfully run 0 times, so let it do so. – Pugilism

@Pugilism I have an else – Eldreeda

Answer 1

Since sys.argv is a list, you can use slicing sys.argv[1:]:

def main(argv):
    for x in argv[1:]:
        build(x)

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

But, if you can only have one script parameter, just get it by index: sys.argv[1]. But, you should check if the length of sys.argv is more than 1 and throw an error if it doesn't, for example:

def main(argv):
    if len(argv) == 1:
        print "Not enough arguments"
        return
    else:
        build(argv[1])

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

Token

Answer 2

The real answer is to learn about and use argparse, though.

Publication

I've used that before, what is the benefit between it and sys.argv or using optparse vs something else – Eldreeda

It's the thing to use for arg parsing (unless you need cliff). It has a great API. It's not deprecated like optparse. It standardizes the look+feel of your script. – Publication