execute code only while sys.argv[1] exist

I have this code:

#!/usr/bin/python

import os.path
import sys
if len(sys.argv)<2:
   print"You need to specify file!" 
if (os.path.isfile(sys.argv[1])):
   print "File <%s> exist" % sys.argv[1]
elif (sys.argv[1] == "--help"):
  print "Add (only)one file argument to command"
  print "--help                   print this screen"
  print "--autor                  autor name and email adress"
  print "--about                  about this program"
elif (sys.argv[1] == "--about"):
  print"Program to identify if the file exists"
  print"Copyright Vojtech Horanek 2015"
elif (sys.argv[1] == "--autor"):
  print"Vojtech Horanek <[e-mail]>"
else:
  print"No file <%s> found" % sys.argv[1]

and i want execute this piece of code only while sys.argv[1] is exist:

if (os.path.isfile(sys.argv[1])):
   print "File <%s> exist" % sys.argv[1]
elif (sys.argv[1] == "--help"):
   print "Add (only)one file argument to command"
   print "--help                   print this screen"  
   print "--autor                  autor name and email adress"
   print "--about                  about this program"
elif (sys.argv[1] == "--about"):
   print"Program to identify if the file exists"
   print"Copyright Vojtech Horanek 2015"
elif (sys.argv[1] == "--autor"):
   print"Vojtech Horanek <[e-mail]>"
else:
  print"No file <%s> found" % sys.argv[1]

if i only start the program without an arguments (python program.py) it’s print this text:

You need to specify file!
Traceback (most recent call last):
File "program.py", line 7, in <module>
if (os.path.isfile(sys.argv[1])):
IndexError: list index out of range

I tried “if sys.argv == 1” but doesnt worked.

Any solutions? Thanks

Answer

if len(sys.argv)<2:
   print"You need to specify file!" 
   sys.exit()

Now your program will terminate entirely if the user didn’t supply any arguments, rather than continuing and raising an exception.