How do I break the while loop with user input

Question

I'm finding it difficult to break a while loop when a user inputs an 'N', it only accepts a 'Y'. Inputting an 'N' will still activate the second while loop.

How do I write such that once user inputs N, the while loop breaks? Thank you in advance :)

choice = 'Y'
while choice == 'Y':

    choice = input('Would you like to continue? (y/n): ').upper()
    # it breaks at this point  

    while (choice != 'Y') or (choice != 'N'):
        choice = input('Please choose Y for yes, or N for no!: ').upper()

Does this answer your question? [Asking the user for input until they give a valid response](https://stackoverflow.com/questions/23294658/asking-the-user-for-input-until-they-give-a-valid-response) — wjandrea, May 23 '20 at 16:32
`(choice != 'Y') or (choice != 'N')` is always true. Any value of `choice` is either not equal to `'Y'` or not equal to `'N'`. You mean `and` instead of `or`. — khelwood, May 23 '20 at 16:37

score 1 · Accepted Answer · answered May 23 '20 at 16:33

1

choice = 'Y'
while choice == 'Y':
    choice = input('Would you like to continue? (y/n): ').upper()
    while (choice != 'Y') and (choice != 'N'): # and, not or!
        choice = input('Please choose Y for yes, or N for no!: ').upper()

answered May 23 '20 at 16:33

Błotosmętek

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oh thank you so much, its such a simple solution. I've only started self learning a week ago, thank you @Blotosmetek for your help – Luqman Buang May 23 '20 at 16:39

Brett La Pierre · Answer 2 · 2020-05-23T18:22:16.110

The other answer provided was fantastic but I want to give an example using the "break" statement.

  while True:
        user_input = input("Run more y/n?")
        if user_input == 'y':
            print("I'm still running in the loop")
        elif user_input == 'n':
            print("I am no longer running and have broken from the while loop.")
            break
        else:
            print("The input is invalid but I'm still running in the loop. Try again")

How do I break the while loop with user input

2 Answers2