Python username and password creator

Question

I am creating a program on python that allows you to create a username and password and here's what I have so far:

#Code
username = str(input("Please enter a username:"))
print("Your username is",username,",proceed?")
raw_input = input()
if raw_input() == 'no':
    re_u = input("Please re-enter username")
else:
    import getpass
    mypass = getpass.getpass("Please enter your password:")

And the trouble i'm receiving is: Traceback (most recent call last): line 6, in if raw_input() == 'no': TypeError: 'str' object is not callable.

Please help

are you writing this in python 2.7 or 3.x?

Findlay Smith
– Findlay Smith

2016-11-12 21:24:22 +00:00
Commented Nov 12, 2016 at 21:24 — Findlay Smith
– Findlay Smith, Commented Nov 12, 2016 at 21:24

Ahasanul Haque · Accepted Answer · 2016-11-12 21:48:46Z

1

In the Line:

raw_input = input()

If you are using python 2.7, You are overriding the default raw_input and making it a string.

If you are using python 3, you are creating a string raw_string with the value of input()

So, in both cases, when you try to call raw_input like a function, you will receive the error.

TypeError: 'str' object is not callable.

if raw_input() == 'no':

Here, raw_input is a string, not a callable. You can not call it like a function.

Now, so the working code should be like below: (A python 2.7 solution)

import getpass

username = raw_input("Please enter a username:")
username_confirmation = raw_input("Your username is " + username + ",proceed?")
if username_confirmation == "No":
    username = raw_input("Please re-enter username")
else:
    password = getpass.getpass("Please enter your password:")

# do something with username and password

edited Nov 12, 2016 at 21:48

answered Nov 12, 2016 at 21:27

Ahasanul Haque

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3 Comments

Makoto Over a year ago

If this is Python 3, then raw_input isn't a default method anymore, since it was renamed to input instead.

Ahasanul Haque Over a year ago

@Makoto I got the feeling that this is python 2, because OP is surely a beginner and using raw_input() wrongly. If this is python 3, he shouldn't be familiar with it. Anyway, I have edited the answer for both the cases.

Makoto Over a year ago

print became a function in Python 3, and input behaves the same as raw_input did in Python 2 (as in Python 2, input did something distinct too). I strongly doubt that the OP is using Python 2, but thanks for correcting it.

matias elgart · Accepted Answer · 2016-11-12 21:25:11Z

0

raw_input is a string, but you're trying to use it as a function with the () at the end of it. use this instead:

if raw_input == 'no':

answered Nov 12, 2016 at 21:25

matias elgart

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