How to exit a while True loop using a different format condition in Python?

Question

I found the following code example online, which exits a while True loop by using a specific integer:

while True:
i = int(input('Please enter an integer (or 0 to exit):\n'))
if i == 0:
    print("Exiting the Program")
    break
print(f'{i} square is {i ** 2}')

I would like to know how to exit the loop using a str input (like "exit"), instead of a specific int (0 in this example). I messed around by creating one input per type:

while True:
    i_str = input('Please enter an integer ( or "exit" to exit): ')
    if i_str == "exit":
        print("Exiting the Program")
        break
    i_int = int(input('Please enter an integer (or "exit" to exit): '))
    print(f'{i_int} square is {i_int ** 2}')

...but it prints out smth every other try, and throws a ValueError when computing 'exit' on the wrong try.

I basically can't figure out how to implement this properly, neither am I sure that I should even be using two different input formats to solve this.

Krishna Chaurasia · Accepted Answer · 2021-02-16 10:27:50Z

2

You can use if-else and convert to int if it is not exit as:

while True:
    i_str = input('Please enter an integer ( or "exit" to exit): ')
    if i_str == "exit":
        print("Exiting the Program")
        break
    else:
        i_int = int(i_str)
        print(f'{i_int} square is {i_int ** 2}')

answered Feb 16, 2021 at 10:27

Krishna Chaurasia

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2 Comments

Krishna Chaurasia Over a year ago

Glad to be of help, @Effojire, please consider accepting the answer if it resolved your problem.

Effojire Over a year ago

Upvoted and accepted. Upvote won't show as I'm a newbie here. Thanks for letting me know abt the accepting option, had no idea there was such a thing.

Michiel · Accepted Answer · 2021-02-16 10:59:13Z

0

you can just cast it to a string first and validate it then cast it to int and then put that all in a try-catch to catch any exceptions you can use the example below

while True:
try:
    i = str(input('Please enter an integer (or 0 to exit):\n'))
    if "exit" in i:
        print("Exiting the Program")
        break
    else:
        i = int(i)
    if i == 0:
        print("Exiting the Program")
        break
    print(f'{i} square is {i ** 2}')
except:
    print("Exiting the Program")

answered Feb 16, 2021 at 10:59

Michiel

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Comments

Prudhvi · Accepted Answer · 2021-02-16 11:42:00Z

0

You can just use try-except, program will raise an exception if it is unable to convert the entered number to integer.

while True:
    try:
        i = int(input('Please enter an integer ( or "exit" to exit): '))
        print(f'{i} square is {i ** 2}')
    except:
        print("Exiting the Program")
        break

answered Feb 16, 2021 at 11:42

Prudhvi

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