Seeing if a user typed in an integer for input

Question

hours = 0.0
while hours != int:
    try:
        hours = int(raw_input('Enter how many hours ahead you want to know the weather: '))
        break
    except ValueError:
        print "Invalid input. Please enter an integer value."
        hours = int(raw_input('Enter how many hours ahead you want to know the weather: '))

Okay, so I am trying to make sure a user inputs an integer value. If I don't type in an integer the "Invalid input, Please enter an integer value." will come up and I will then enter in another non integer value and will end up getting an error message. So why does it work the first time and not the second?

score 4 · Answer 1 · answered Dec 16 '14 at 01:42

4

Use break statement to get out of the loop when user input correct integer string.

hours = 0.0
while True:
    try:
        hours = int(raw_input('Enter how many hours ahead you want to know the weather: '))
        break  # <---
    except ValueError:
        print "Invalid input. Please enter an integer value."
        # No need to get input here. If you don't `break`, `.. raw_input ..` will
        #   be executed again.

BTW, hours != int will always yield True. If you want to check type of the object you can use isinstance. But as you can see in the above code, you don't need it.

answered Dec 16 '14 at 01:42

falsetru

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How do I fix hours != int? – John Emaus Dec 16 '14 at 01:45
@JohnEmaus, `isinstance(hours, int)`. But that does not help here. `while` loop with `break` will solve your problem without checking type. – falsetru Dec 16 '14 at 01:47
@JohnEmaus: You don't need `hours != int`. The answer here uses `while True:`. – Greg Hewgill Dec 16 '14 at 01:47

Seeing if a user typed in an integer for input

1 Answers1