Python if being ignored

Question

It is skipping the if Ctype == 1 (2, 3, etc.) down at the bottom, no idea why.

print("Starting game.")
print("Game Started.")
print('\n')
CharName = input('Create a character name: ')
print('\n' * 20)
print('Class Selection: (Ensure it as written as you read it)')
print('\n')
print('Classes:')
print('[1] Heavy')
print('[2] LightFooted')
print('[3] Warrior')
print('\n')
print('To Come:')
print('Stealth')
print('\n')
Ctype = input('Choose class: ')
if Ctype == 1:
    print('test')
if Ctype == 2:
    print('test')
if Ctype == 3:
    print('test')
if Ctype == 4:
    print('Try again... another time.')
    time.wait(2)

`input()` returns a string, you compare to integer, convert one of those — PRMoureu, Sep 09 '17 at 09:45
your code would work in `python 2.7+` but wont in `python 3+` without some tweaking — Bijoy, Sep 09 '17 at 09:49

score 2 · Accepted Answer · answered Sep 09 '17 at 09:47

2

input() returns a string in Python3.

So Ctype is a string. You then compare with an integer, thus the equality check fails.

Convert the string to an int, like this:

Ctype = int(Ctype)

answered Sep 09 '17 at 09:47

gsamaras

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Will try it now. Yep, makes sense and works, will keep it in mind for future. Appreciated. – TGBF Project Sep 09 '17 at 09:50
Good @TGBFProject, accept the answer then! =) – gsamaras Sep 09 '17 at 09:50

Python if being ignored

1 Answers1