My python condition checking the input from user is a negative number doesn't work

Question

I write a code in python to ask a user to input a number and check if is it less than or equal to zero and print a sentence then quit or otherwise continue if the number is positive , also it checks if the input is a number or not

here is my code

top_of_range = input("Type a number: ")

if top_of_range.isdigit():
    top_of_range = int(top_of_range)

    if top_of_range <= 0:
        print ("please enter a nubmer greater than 0 next time.")
        quit()

else:
    print(" please type a number next time.")
    quit()

Why do you think that it doesn't work?

mkrieger1
– mkrieger1

2022-09-22 22:14:18 +00:00
Commented Sep 22, 2022 at 22:14 — mkrieger1
– mkrieger1, Commented Sep 22, 2022 at 22:14
I don't think - counts as a digit.

Mark Ransom
– Mark Ransom

2022-09-22 22:17:53 +00:00
Commented Sep 22, 2022 at 22:17 — Mark Ransom
– Mark Ransom, Commented Sep 22, 2022 at 22:17

Omer Khalil · Accepted Answer · 2022-09-26 12:17:28Z

1

As the help on the isdigit method descriptor says:

Return True if the string is a digit string, False otherwise. A string is a digit string if all characters in the string are digits and there is at least one character in the string.

When you pass a negative number, e.g. -12. All characters in it are not a digit. That is why you're not getting the desired behavior.

You can try something like the following:

top_of_range = input("Type a number: ")

if top_of_range.isdigit() or (top_of_range.startswith("-") and top_of_range[1:].isdigit()):
    top_of_range = int(top_of_range)

    if top_of_range <= 0:
        print ("please enter a nubmer greater than 0 next time.")
        quit()

else:
    print(" please type a number next time.")
    quit()

Based on Mark's comment, if we employ the EAFP principle it will look something like this:

top_of_range = input("Type a number: ")

try:
    top_of_range = int(top_of_range)

    if top_of_range <= 0:
        print ("please enter a nubmer greater than 0 next time.")
        quit()
except ValueError:
    print(" please type a number next time.")
    quit()

edited Sep 26, 2022 at 12:17

answered Sep 22, 2022 at 22:17

Omer Khalil

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1 Comment

Mark Ransom Over a year ago

In Python it's considered better to just do the conversion and detect an error with try/except rather than qualify the input first. It's called the EAFP principle.

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