Python command line arguments linux

Question

I have this little program(I know there is a lot of errors):

#!/usr/bin/python

import os.path
import sys

filearg = sys.argv[0]

if (filearg == ""):
    filearg = input("")

else:

    if (os.path.isfile(filearg)):
        print "File exist"

    else:
        print"No file"
        print filearg
        print "wasn't found"

If i start it by typing python file.py testfile.txt

the output will be always(even if the file doesn't exist):

File exist

If you don't know what iam want from this program, i want to print "File 'filename' wasn't found" if the file isn't exist and if it's exist iam wan't to print "File exist"

Any ideas to solve it? Thanks

Too lazy to debug? print filearg...

Karoly Horvath
– Karoly Horvath

2015-05-27 12:16:56 +00:00
Commented May 27, 2015 at 12:16 — Karoly Horvath
– Karoly Horvath, Commented May 27, 2015 at 12:16

Padraic Cunningham · Accepted Answer · 2015-05-27 12:16:13Z

11

It should be sys.argv[1] not sys.argv[0]:

filearg = sys.argv[1]

From the docs:

The list of command line arguments passed to a Python script. argv[0] is the script name (it is operating system dependent whether this is a full pathname or not). If the command was executed using the -c command line option to the interpreter, argv[0] is set to the string '-c'. If no script name was passed to the Python interpreter, argv[0] is the empty string.

edited May 27, 2015 at 12:16

answered May 27, 2015 at 12:14

Padraic Cunningham

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1 Comment

chown Over a year ago

this is correct. sys.argv[0] will always exist because it is the file that is currently being executed (the python script). The first argument always starts at sys.argv[1].

Leon · Accepted Answer · 2015-05-27 12:15:54Z

4

The first argument is always the name of the file being executed. This is true for a number of programming languages, you need to use sys.argv[1]

answered May 27, 2015 at 12:15

Leon

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