How to make a loop in python three that stops when you input the correct word/number

Question

I am making a text based adventure game in python 3 and I was wondering what the simplest loop is. Using the code I have, it continues to print "whats the number" even when you put the correct number, also giving 9 as input doesnt work. It also doesn't work when I give ("8","9") . Here is my code :

print("whats the number?")
required_number = ("8" or "9")

while True:
    number = input()
    if number == required_number:
        print ("GOT IT")
    else: print ("Wrong number try again")

Answer 1

Try this :

print("whats the number?")
required_number = [8,9]
while True:
    number = int(input())
    if number in required_number :
        print('GOT IT')
        break
    else:
        print('Wrong number try again')

Sample output in shell :

whats the number?
5
Wrong number try again
2
Wrong number try again
4
Wrong number try again
8
GOT IT

Answer 2

print("whats the number?")

required_number = [8,9]

while True: number = input()

if number in required_number:

    print ("GOT IT")
    break

else:
    print ("Wrong number try again")

Answer 3

The word you are looking for in place of == is in since required_number is a tuple you're looking to see if the input is in required_number . Also the correct syntax for the tuple would be using a comma not or .

I also would make required_number plural to be a more accurate description of what it holds, and you probably want to use integers and not strings.

required_numbers = (8, 9)

while True:
    number = int(input("whats the number?"))
    if number in required_numbers:
        print("GOT IT")
        break #Stop asking
    else:
        print ("Wrong number try again")

Answer 4

1. input treats it as a str in Python 3.x

2. Preferably use a list for the required numbers

3. Using in to check for the number in the required_numbers

4. Put it in a try block to catch value error exceptions.

Hence:

required_number = [8,9]  # a list of integer types

while True:
    try:
        number = int(input("whats the number? "))   # Using `int` to convert the `str`
        if number in required_number:
            print ("GOT IT")
            break   # break out when the number is found
        else:
            print ("Wrong number try again")
    except ValueError:
        print("Invalid Input, Please enter an integer only.")

Note: == determines if the values are equal , while in operator iterates over the list of elements and returns True or False .

OUTPUT :

whats the number? g
Invalid Input, Please enter an integer only.
whats the number? abc
Invalid Input, Please enter an integer only.
whats the number? 3
Wrong number try again
whats the number? 9
GOT IT

Answer 5

If your required_number or the input will accommodate a string , then you can use this:

required_number = [8,9]
required_number = str(required_number)
number = None

while True:
    number = input("Write a number: ")
    if number in required_number:
        print ("GOT IT")
    else: 
        print ("Wrong number try again")

Output:

Write a number: 3
Wrong number try again
Write a number: 8
GOT IT
Write a number: Hi
Wrong number try again

Answer 6

Try this Method

print('Enter a Number:')
required_number = ['8','9']

while True:
number = input()
if number in required_number:
    print ("GOT IT")
    break
else:
    print ("Wrong number try again")