Validating user input strings in Python

Question

So I've searched just about every permutation of the words "string", "python", "validate", "user input", and so on, but I've yet to come across a solution that's working for me.

My goal is to prompt the user on whether or not they want to start another transaction using the strings "yes" and "no", and I figured that string comparison would be a fairly easy process in Python, but something just isn't working right. I am using Python 3.X, so input should be taking in a string without using raw input, as far as I understand.

The program will always kick back invalid input, even when entering 'yes' or 'no', but the really weird thing is that every time I enter a a string > 4 characters in length or an int value, it will check it as valid positive input and restart the program. I have not found a way to get valid negative input.

endProgram = 0;
while endProgram != 1:

    #Prompt for a new transaction
    userInput = input("Would you like to start a new transaction?: ");
    userInput = userInput.lower();

    #Validate input
    while userInput in ['yes', 'no']:
        print ("Invalid input. Please try again.")
        userInput = input("Would you like to start a new transaction?: ")
        userInput = userInput.lower()

    if userInput == 'yes':
        endProgram = 0
    if userInput == 'no':
        endProgram = 1

I have also tried

while userInput != 'yes' or userInput != 'no':

I would greatly appreciate not only help with my problem, but if anyone has any additional information on how Python handles strings that would be great.

Sorry in advance if someone else has already asked a question like this, but I did my best to search.

Thanks all!

~Dave

Martijn Pieters · Accepted Answer · 2013-05-19 13:20:57Z

12

You are testing if the user input is yes or no. Add a not:

while userInput not in ['yes', 'no']:

Ever so slightly faster and closer to your intent, use a set:

while userInput not in {'yes', 'no'}:

What you used is userInput in ['yes', 'no'], which is True if userInput is either equal to 'yes' or 'no'.

Next, use a boolean to set endProgram:

endProgram = userInput == 'no'

Because you already verified that userInput is either yes or no, there is no need to test for yes or no again to set your flag variable.

answered May 19, 2013 at 13:20

Martijn Pieters

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3 Comments

user2398870 Over a year ago

Wow. Such a simple mistake. Thanks for the prompt reply. I guess sometimes you just need a second pair of eyes to spot things.

user2398870 Over a year ago

As a side note, could you help me learn why my original method of while userInput != 'yes' or userInput != 'no': wasn't working?

Martijn Pieters Over a year ago

@user2398870: If userInput is set to 'yes', then != 'no' is True. That is not what you wanted to test. :-) Changing the or to and would make that version work.

MrTg · Accepted Answer · 2016-09-11 21:48:14Z

1

def transaction():

    print("Do the transaction here")



def getuserinput():

    userInput = "";
    print("Start")
    while "no" not in userInput:
        #Prompt for a new transaction
        userInput = input("Would you like to start a new transaction?")
        userInput = userInput.lower()
        if "no" not in userInput and "yes" not in userInput:
            print("yes or no please")
        if "yes" in userInput:
            transaction()
    print("Good bye")

#Main program
getuserinput()

answered Sep 11, 2016 at 21:48

MrTg

111 bronze badge

Comments

Pawan Salwe · Accepted Answer · 2022-05-11 20:40:02Z

-2

s=input()
    for i in range(5):
        for r in s:
            if r.isalnum():
                print("True")
                break
            elif r.isalpha():
                print("True")
                break
            elif r.isdigit():
                print("True")
                break
            elif r.islower():
                print("True")
                break
            elif r.isupper():
                print("True")
                break
            else:
                print("False")
                break

edited May 11, 2022 at 20:40

user17242583

answered May 11, 2022 at 10:26

Pawan Salwe

1

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