Exit python program when argument is less than 0

Question

I'd like the program to exit if the input number is less than 0, but sys.exit() isn't doing the trick. This is what I have now:

if len( sys.argv ) > 1:
    number = sys.argv[1]

if number <= 0:
    print "Invalid number! Must be greater than 0"
    sys.exit()

Answer 1

Your test is failing because number is a string.

>>> '-1' <= 0
False

You need to convert number to an integer:

number = int(sys.argv[1])

Note that in Python 3.0 your code would have given an error, allowing you to find your mistake more easily:

>>> '-1' <= 0
Traceback (most recent call last):
  File "<pyshell#0>", line 1, in <module>
    '-1' <= 0
TypeError: unorderable types: str() <= int()