Python: small while loop error

Question

Im new to python and programming in general and I came across this problem while fiddling around with a simple while loop. The loop takes input to evaluate two possible passwords:

    print('Enter password')
    passEntry = input()

    while passEntry !='juice' or 'juice2':
      print('Access Denied')
      passEntry = input()
      print(passEntry)

    print('Access Granted')

It doesnt seem to be accepting juice or juice2 as valid.

Also just accepting one password like:

    while passEntry != 'juice' :

will not work, while:

    while passEntry !='juice' :

works fine. I cant seem to find the reason for these issues (Only difference between the latter two is the space after the =). Any help is appreciated.

This may not reason of error I guess!

Grijesh Chauhan
– Grijesh Chauhan

2012-12-28 05:49:06 +00:00
Commented Dec 28, 2012 at 5:49 — Grijesh Chauhan
– Grijesh Chauhan, Commented Dec 28, 2012 at 5:49

Emil Sit · Accepted Answer · 2012-12-28 05:59:06Z

7

First, you should use Python's getpass module to get a password portably. For example:

import getpass
passEntry = getpass.getpass("Enter password")

Then, the code you have written to guard the while loop:

while passEntry != 'juice' or 'juice2':

gets interpreted by the Python interpreter as a while loop with the guard expression

(passEntry != 'juice') or 'juice2'

This is always true because regardless of whether passEntry equals "juice" or not, "juice2" will be considered as true when interpreted as a boolean.

In Python, the best way to test membership is to use the in operator, which works for a variety of data types such as a list or a set or a tuple. For example, a list:

while passEntry not in ['juice', 'juice2']:

edited Dec 28, 2012 at 5:59

answered Dec 28, 2012 at 5:53

Emil Sit

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Comments

avasal · Accepted Answer · 2012-12-28 05:49:19Z

3

you can use

while passEntry not in ['juice' ,'juice2']:

answered Dec 28, 2012 at 5:49

avasal

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Comments

Vlad · Accepted Answer · 2012-12-28 05:49:14Z

1

How about:

while passEntry !='juice' and passEntry!= 'juice2':

and using raw_input() instead of input()?

input() evaluates the input as if it were Python code.

answered Dec 28, 2012 at 5:49

Vlad

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1 Comment

Baker644 Over a year ago

Yes this was it for the first part, made logical error using or instead of and. Thanks

BrenBarn · Accepted Answer · 2012-12-28 05:49:46Z

1

passEntry !='juice' or 'juice2' means (pass != 'juice') or ('juice2'). "juice2" is a nonempty string, so it is always true. Thus your condition is always true.

You want to do passEntry != 'juice' and passEntry != 'juice2', or more nicely passEntry not in ('juice', 'juice2').

answered Dec 28, 2012 at 5:49

BrenBarn

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Comments

jackcogdill · Accepted Answer · 2012-12-28 05:47:22Z

0

Does this work?

while passEntry !='juice' and passEntry !='juice2':

answered Dec 28, 2012 at 5:47

jackcogdill

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Comments

HelloUni · Accepted Answer · 2012-12-28 09:53:31Z

0

Your error is in the way you wrote the while statement.

while passEntry !='juice' or 'juice2':

That line will always be true, when read by the python interpreter. And also instead of:

passEntry = input()

Use:

passEntry = raw_input()

(Unless you are using Python 3)

the input in Python 2 evals your input.

This will be the proper code:

print('Enter password')
passEntry = raw_input()

while passEntry != 'juice' and passEntry != 'juice2':
    print('Access Denied')
    passEntry = raw_input()
    print(passEntry)

print('Access Granted')

answered Dec 28, 2012 at 9:53

HelloUni

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1 Comment

Baker644 Over a year ago

Ok, im running python3, so i guess thats why raw_input() gave me problems. Thanks

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