Using sys.argv to implement command line arguments?

Question

I'm fairly confused as to the proper implementation of command line arguments in a program. I've looked around and am confused whether or not a for loop is necessary. I know for sure that there will always be only two arguments passed to this program. Ultimately, is a for loop necessary in this situation? Here's the code in the main function:

def main():
try:
    if len(sys.argv) > 1:
        filename = sys.argv[1]
        length = int(sys.argv[2])
    wordDict = readFile(filename)
except IOError:
    print("Error: file not found.")

Thank you all for your input!

mpursuit · Accepted Answer · 2014-10-27 21:17:21Z

2

You do not need to use a loop if you know exactly how many arguments you want.

I have noticed that in your code that you check the number of sys.argv is greater than 1, you need to check that it is greater than 2 as the first sys.argv (sys.argv[0]) is the name script that you are running https://docs.python.org/3/library/sys.html?highlight=sys.argv#sys.argv

If only one argument is passed to your program it will currently result in an IndexError.

answered Oct 27, 2014 at 21:17

mpursuit

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1 Comment

tdelaney Over a year ago

Also, if len(sys.argv) is less than 2, it will raise a NameError.

Ragajo · Accepted Answer · 2014-10-27 21:01:11Z

2

you can try argparse: https://docs.python.org/2/library/argparse.html

import argparse


parser = argparse.ArgumentParser(description='What ever the purpose of the script is')

parser.add_argument('filename', type=str,
                   help='path to the file')

parser.add_argument('length', type=int,
                   help='some length')

args = parser.parse_args()
print args.filename
print args.length

answered Oct 27, 2014 at 21:01

Ragajo

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