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

• Karoly Horvath: Too lazy to debug? print filearg...

Answer 1

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.

• chown: 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].

Answer 2

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]