我想对int（）进行输入验证，但验证str（）

Question

I want the code to "break" when the user's input is <=0 AND when the input = "stop". This is what I have so far.

while True:

    try:
        x = input("how many times do you want to flip the coin?: ")
        if int(x) <= 0 or x.lower() == "stop": 
            break
        x = int(x)
        coinFlip(x)


     except ValueError:
        print ()
        print ("Please read the instructions carefully and try one more time! :)")
        print ()

I get the error:

    if int(x) <= 0 or str(x).lower() == "stop":
ValueError: invalid literal for int() with base 10: 'stop'

Answer 1

you get the exception because the first condition that's evaluated is int(x) <= 0 and x is not really an integer at this point.

you can change the order of the the conditions:

if x.lower() == 'stop' or int(x) <=0

this way you check for 'stop' first, and don't evaluated int(x) (because the the or condition already evaluates to True ). any string that is not an integer and not 'stop' would cause the ValueError exception which you're already handling.

Answer 2

You get a ValueError because you can't convert the string 'stop' to an integer.

One way to solve this would be to use a helper method that correctly catches the ValueError and then checks if the string is stop :

def should_stop(value):
    try:
        return int(value) <= 0
    except ValueError:
        return value.lower() == "stop"

while True:
    x = input("how many times do you want to flip the coin?: ")
    if should_stop(x): 
        break