python在程序中传递命令行参数

Question

I have below python program takes argument from command line.

import argparse
import sys

choice=['orange','apple', 'grapes', 'banna']

def select():
    parser = argparse.ArgumentParser(prog='one', description='name test')
    parser.add_argument(
        'fname',
        action="store",
        type=str,
        choices=choice,
        help="furits name")

    args = parser.parse_args(sys.argv[1:2])

    print 'selected name {0}\n'.format(args.fname)

if __name__ == '__main__':
    select()

this works

 python s.py apple
selected name apple

How can inline argument with-in main function. I tried this but its not working.

change main line this.

if __name__ == '__main__':
    sys.argv[0]='apple'
    select()

getting below error.

 usage: one [-h] {orange,apple,grapes,banna} one: error: too few arguments 

How can I achieve this in argument?

thanks -SR

Answer 1

Your index is wrong sys.argv[0] will be the path of the python script. What you want is:

if __name__ == '__main__':
    if len(sys.argv) == 1:
        sys.argv.append("apple")
    select()

But, this is a weird way of doing things. After a bit more thought, this occurred to me:

choice=['orange','apple', 'grapes', 'banna']

def select():
    parser = argparse.ArgumentParser(prog='one', description='name test')
    parser.add_argument(
        'fname',
        nargs='?',
        default='apple',
        action="store",
        type=str,
        choices=choice,
        help="furits name")

    args = parser.parse_args(sys.argv[1:2])

    print 'selected name {0}\n'.format(args.fname)

if __name__ == '__main__':
    select()

Note the nargs='?' and default='apple' additions to the call to add_argument(). These make the parameter optional and set the default value to "apple" if no argument is supplied.