Question

So I am doing some of the suggested readings here: http://tinyurl.com/5qv3yb and am trying to do an exercise about writing a password asking program. I was successful, but with one thing I can't figure out. When it asks for my password I have to put quotation marks " " around it or it says "name is X is not defined". How do I make it not need the quotations?

Answer 1

Note: I am using TextMate

Answer 2

Can you link your code? It shouldn't be a problem of the text editor, at least, hopefully not.

Answer 3

password = 'mozart'
question = "What is your password? "
print "Please enter your password to log in"

x = input(question)

while x != password:
if x != password:
print "Password Incorrect: 2 tries remaining"
x = input(question)
if x!= password:
print "Password Incorrect: 1 try remaining"
x = input(question)
if x!= password:
print "Password Incorrect. You have been denied access"
quit()
else: x == password
print "You have succesfully logged in"

Answer 4

Try using codepad or something like that to post your code afterwards. It is easier to see and debug. Did you try raw_input instead of only input? raw_input returns a string.

Answer 5

I think you'll need a variable that iterates from three to zero as your unsuccessful tries happen, rather than nested ifs

Answer 6

pully6, yeah, that would be a lot better, but I wasn't sure how to do it, so I kind of just improvised something else that worked. BMP, thanks so much, that was what was wrong! I knew it was something simple. I'm still learning the basics here, haha.

Answer 7

No worries, I'm the same just learning basics. I'm working on a solution to your prob, starting with something like:

for triesRemaining in range(0,3):
raw_input(question) != password

Answer 8

Also, thanks for letting me know about codepad BMP. Good to know :).

Answer 9

Do you mind putting that code into context of the problem pully6? I don't understand where to plug it in.

here is a link using codepad to what I have: http://codepad.org/SlDIF9Bm

Answer 10

I've re-written your problem, it works now. I surprised myself actually! Used an iterating variable triesRemaining to give the user 3 chances. http://codepad.org/Nxd7yYia

Answer 11

Other cleaner option is to set a boolean value as false and then, if the input matches the password, this boolean becomes True, exiting the loop. Something like while not isPass: #code if input == password: isPass = True.
And, no problem mate, the beginning is harsh but it is worth it. And codepad is just to make easier to help you out / clearer to see the code.

Answer 12

ah boolean is a good idea.

I found the learning curve for the first few lectures quite steep but now I'm up to lect 5 I'm making good progress. stick with it :)

Answer 13

Thanks a bunch for the help guys :). OpenStudy and the people on it make this process a lot easier.

Answer 14

raw_input() asks the user for some text and returns it as a string. That's what you want to use (almost always).

input() asks the user for some text, then tries to interpret it as python code. That is dangerous. A malicious user can inject any arbitrary code into your program that way. Also, that's why it complained about not finding the name. If the user types a string with no quotes around it, they're really typing a variable name. If they happen to type the name of a variable you're using, then there won't be an error. If what they type isn't a variable name, then python doesn't know what they're trying to do, so it complains.