ValueError in python exception: how to handle while user input is nested in argument of function. Want to return message when exception occurs

Question

My problem here is that if the user inputs a value that is not an integer, it throws a ValueError: invalid literal for int() with base 10:'example' i want to handle this exception if the user inputs a string or float based value and return print("Sorry thats not right, try again") and replay the function. how can this be accomplished? I fear it is because the input function is posed as an argument inside the configselect() function that this might not be possible?

choice = [1, 2, 3]


def configselect(choice):
    if choice == 1:
        print("you chose 962")
    elif choice == 2:
        print("you chose Audience")
    elif choice == 3:
        print("you chose mAP")
    else:
        print('Thats not a valid Number. Please try Again.')


configselect(int(input("Are you configuring a (1)962, (2)Audience, or (3)mAP? ")))

Of course, you need to check if is it a number or not before you pass the argument. int() cannot convert letters or symbols the integer — user15801675
– user15801675, Commented Jul 28, 2021 at 1:48
You catch it like any other exception. Do you know what exceptions are? Do you know what try ... except is? — Silvio Mayolo
– Silvio Mayolo, Commented Jul 28, 2021 at 1:49
Read 1. docs.python.org/3/tutorial/errors.html#handling-exceptions or 2. docs.python.org/3/library/stdtypes.html#str.isdigit — Niel Godfrey P. Ponciano
– Niel Godfrey P. Ponciano, Commented Jul 28, 2021 at 1:51

user15801675 · Accepted Answer · 2021-07-28 01:52:15Z

1

You should use try...except ValueError, as a ValueError is raised when integer cannot convert something to an integer

choice = [1, 2, 3]


def configselect(choice):
    if choice == 1:
        print("you chose 962")
    elif choice == 2:
        print("you chose Audience")
    elif choice == 3:
        print("you chose mAP")
    
while True:
    try:
        val=int(input("Are you configuring a (1)962, (2)Audience, or (3)mAP? "))
        if val in choice:
            configselect(val)
            break
        else:
            print("Please select either 1,2 or 3")
    except ValueError:
        print('Thats not a valid Number. Please try Again.')

answered Jul 28, 2021 at 1:52

user15801675

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1 Comment

Crowfeatherjoe Over a year ago

this is exactly what i was looking for, thank you so much, earlier i had tried the try/except inside of the function and my issue was that the value was hitting the exception before it ever made it to the try/except. cheers

CyanoKobalamyne · Accepted Answer · 2021-07-28 01:50:47Z

0

If you want to catch the exception, you need to add a try-except block:

try:
    configselect(int(input("Are you configuring a (1)962, (2)Audience, or (3)mAP? ")))
except ValueError:
    print("Sorry, that's not right, try again")

If you want to prompt for input continuously until the user enters something acceptable, you need to wrap this in a loop:

while True:
    try:
        configselect(int(input("Are you configuring a (1)962, (2)Audience, or (3)mAP? ")))
    except ValueError:
        print("Sorry, that's not right, try again")
    else:
        break

answered Jul 28, 2021 at 1:50

CyanoKobalamyne

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