def check_range(myrange): if range(myrange) != range(10,89): return "False" else: return "True"
For this to work out, the user's input would have to be a giant string containing 10, 11, 12, 13, etc.
Not quite, actually.
Presuming that myrange is an integer, range(myrange) will generate a list of integers starting with 0 and continuing up to (but not including) myrange, while range(10,89) generates a list of integers starting with 10 and continuing up to (but not including) 89. These lists can never be equal, because range(10,89) will never include the integers 0-9 but range(myrange) will (unless myrange is less than 10).
Unless I mistunderstood your requirements, what you're probably looking for is:
if myrange in range(10,90): # "in" is the key word here return True else return False
This is, however, the correct solution. :) Presuming again, of course, that myrange is an integer -- but be aware that user input normally comes in the form of strings, so it will be necessary, at some point, to create an integer from that string using int().
Jeff Shannon Technician/Programmer Credit International
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