New to python. Need help to understand code

Question

Hello I am new to python just started . Would be glad if someone can help me to understand this piece of code.

#!/usr/bin/python

import sys, getopt

def main(argv):
   inputfile = ''
   outputfile = ''
   try:
      opts, args = getopt.getopt(argv,"hi:o:",["ifile=","ofile="])
   except getopt.GetoptError:
      print 'test.py -i <inputfile> -o <outputfile>'
      sys.exit(2)
   for opt, arg in opts:
      if opt == '-h':
         print 'test.py -i <inputfile> -o <outputfile>'
         sys.exit()
      elif opt in ("-i", "--ifile"):
         inputfile = arg
      elif opt in ("-o", "--ofile"):
         outputfile = arg
   print 'Input file is "', inputfile
   print 'Output file is "', outputfile

if __name__ == "__main__":
   main(sys.argv[1:])

Answer 1

To the best of my ability, please find my comments ( # comments ) below:

#!/usr/bin/python

# importing libraries    
import sys, getopt

def main(argv):
   # defining empty string variables
   inputfile = ''
   outputfile = ''

   # try/except block. Try this code, if it fails, handle it
   try:
      # looks like you are parsing command line arguments
      opts, args = getopt.getopt(argv,"hi:o:",["ifile=","ofile="])
   except getopt.GetoptError:
      print 'test.py -i <inputfile> -o <outputfile>'
      sys.exit(2)
   #iterate through the cmd line arguments
   for opt, arg in opts:
      # of option is -h, do this (same with the following)
      if opt == '-h':
         print 'test.py -i <inputfile> -o <outputfile>'
         sys.exit()
      elif opt in ("-i", "--ifile"):
         inputfile = arg
      elif opt in ("-o", "--ofile"):
         outputfile = arg
   print 'Input file is "', inputfile
   print 'Output file is "', outputfile

# python main function
if __name__ == "__main__":
   # call method main and pass the cmd line arguments
   main(sys.argv[1:])