On Tue, Jun 7, 2011 at 11:26 PM, Matthew Brunt <mbrun...@gmail.com> wrote:
> i'm very new to python (currently going through a python for beginners
> book at work to pass the time), and i'm having trouble with an if
> statement exercise. basically, i'm creating a very simple password
> program that displays "Access Granted" if the if statement is true.
> the problem i'm having is that no matter what i set the password to,
> it seems like it's ignoring the if statement (and failing to print
> "Access Granted"). the code is copied below. i'm pretty sure it's my
> own programming ignorance, but i would greatly appreciate any
> feedback.
Actually, no, this case it's not your ignorance. What is probably
going on is that you are using Python 2, but the book is going with
Python 3. I won't get into the details, but the function that does
what input() does in Python 3, is called raw_input() in Python 2 (and
the Python 2 input() function is advised against using for security
reasons). Probably if you change input to raw_input in your program,
it does do what you want it to do.
> # Granted or Denied
> # Demonstrates an else clause
>
> print("Welcome to System Security Inc.")
> print("-- where security is our middle name\n")
>
> password = input("Enter your password: ")
>
> if password == "a":
> print("Access Granted")
>
> input("\n\nPress the enter key to exit.")
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--
André Engels, andreeng...@gmail.com
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