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Hi guys.. i found something strange on the behavior of OptionParser
If I have this sample code : 
import sys
from optparse import OptionParser
if __name__ == '__main__':
 parser = OptionParser()
 parser.add_option("-p", "--p", help="The P of python", default=None)
 parser.add_option("-y", "--y", help="The Y of python", default=None)
 
 (options,args) = parser.parse_args(sys.argv)
 print options
and i execute :
myFile.py -p -y
the options.p will be -y and options.y will be None
worst.. if my user say :
myFile.py -p -y 'thon'
options.p will be -y and options.y will be None...
In all case I think that, if i do 
myFile.py -p -y 
options -p and options -y must be None
and if i want to put -y in options.p i say something like 
myFile.py -p "-y" so i can do after myFile.py -p "-y" -y "thon".

This is not a bug. By default, options in optparse take an argument --
therefore, the -y is taken as the argument to the -p option.
Use e.g. add_option(..., action='store_true') to specify an option that
doesn't take an argument.