How to handle specific integer exception in python with try function. How to handle multiple except with try function

Question

This code should tell exactly what error the user is making and prompt to try again.

How can I make custom error messages for each error?

Would there be a much more simpler solution like the do-while in c-programming?

while True:
    height = int(input("Height: "))
    try:
        check_answer = int(height)
        assert (int(height) > 0)
        assert (int(height) < 9)
        break
    except ValueError:
        print("must enter a number")
    except (???):
        print("enter a number greater than 0")
    except (???):
        print("enter a number smaller than 9")

Must you use try-except blocks for the greater than & less than conditions? Can't those just be 'if' statements after the ValueError exception? — Ben
– Ben, Commented Apr 1, 2020 at 16:03
Both asserts would produce exactly the same class of exception if they fail, and that's the only thing you can base an except clause on. — jasonharper
– jasonharper, Commented Apr 1, 2020 at 16:05

blhsing · Accepted Answer · 2020-04-01 16:13:20Z

4

If you must use the assert statement, you can pass the message as a second argument so that it becomes the message of the AssertionError exception:

while True:
    try:
        height = int(input("Height: "))
        assert height > 0, "enter a number greater than 0"
        assert height < 9, "enter a number smaller than 9"
        break
    except ValueError:
        print("must enter a number")
    except AssertionError as e:
        print(str(e))

But what you want to achieve is usually done with simple if statements instead:

while True:
    try:
        height = int(input("Height: "))
    except ValueError:
        print("must enter a number")
    if height <= 0:
        print("enter a number greater than 0")
    elif height >= 9:
        print("enter a number smaller than 9")
    else:
        break

edited Apr 1, 2020 at 16:13

answered Apr 1, 2020 at 16:07

blhsing

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