Python Problems with if statement

Question

I have a problem with this piece of code. Somehow if we put in 'register' or 'Register' as input, it goes to the register function, but after it prints the else: Aswell ("Bad input"); I have made more of these but cant find the fault in this one.

Please help me. Thanks! Here's my code:

def Boot(): 
        print "Type 'Login' + accountname to login."

        x = raw_input("> ")
        Name = x.split()

        if x == "Register" or x == "register":
            print "Registerbla"

        if Name[0] == "Login" or Name[0] == "login":
            print "Loginblabla"
        else:
            print "Bad input"

So what I see after input is: Registerbla Bad input

In editing your post, I notice that you have spaces and tabs in your source. I hope that's just from trying to get it to display properly in stack overflow's renderer -- However, if it isn't, that's a particularly bad thing in python code. You can always run python with python -t script.py to find out if you're mixing spaces and tabs in a bad way. — mgilson
– mgilson, Commented Feb 20, 2013 at 13:59

thegrinner · Accepted Answer · 2013-02-20 14:04:56Z

2

You're missing the else portion of your if statement. Without it, you actually check two separate if statements: the register section and the login/bad input section. Instead, you should use elif:

    if x == "Register" or x == "register":
        print "Registerbla"
    elif Name[0] == "Login" or Name[0] == "login":
        print "Loginblabla"
    else:
        print "Bad input"

Also, consider changing your statements to check against lowercase, like

if x.lower() == "register":
    # Now any capitalized variant of register will work!

edited Feb 20, 2013 at 14:04

answered Feb 20, 2013 at 13:59

thegrinner

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dm03514 · Accepted Answer · 2013-02-20 13:59:01Z

1

when you type in register or Register

if Name[0] == "Login" or Name[0] == "login":

evaluates to false, printing Bad input

answered Feb 20, 2013 at 13:59

dm03514

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