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

user_12948999 0
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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")
try-except python
1 Answers

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
user_68909120
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