If statement always prints (Python)

Question

I'm trying to add an if statement to check for invalid input. It's working like it should and looping back again if the user enters "Yes" and ending if the user enters "No". But for some odd reason no matter what the answer: yes, no, random character, etc. It always prints the "invalid input" statement. I'm trying to make that only print when the answer isn't "Yes" or "No".

while cont == "Yes":
    word=input("Please enter the word you would like to scan for. ") #Asks for word
    capitalized= word.capitalize()  
    lowercase= word.lower()
    accumulator = 0

    print ("\n")
    print ("\n")        #making it pretty
    print ("Searching...")

    fileScan= open(fileName, 'r')  #Opens file

    for line in fileScan.read().split():   #reads a line of the file and stores
        line=line.rstrip("\n")
        if line == capitalized or line == lowercase:
            accumulator += 1
    fileScan.close

    print ("The word", word, "is in the file", accumulator, "times.")

    cont = input ('Type "Yes" to check for another word or \
"No" to quit. ')  #deciding next step
    cont = cont.capitalize()

    if cont != "No" or cont != "Yes":
        print ("Invalid input!")

print ("Thanks for using How Many!")  #ending

Martijn Pieters · Accepted Answer · 2013-07-15 18:28:45Z

8

That's because whatever you enter, at least one of the tests is True:

>>> cont = 'No'
>>> cont != "No" or cont != "Yes"
True
>>> (cont != "No", cont != "Yes")
(False, True)
>>> cont = 'Yes'
>>> cont != "No" or cont != "Yes"
True
>>> (cont != "No", cont != "Yes")
(True, False)

Use and instead:

>>> cont != 'No' and cont != 'Yes'
False
>>> cont = 'Foo'
>>> cont != 'No' and cont != 'Yes'
True

or use a membership test (in):

>>> cont not in {'Yes', 'No'}  # test against a set of possible values
True

answered Jul 15, 2013 at 18:28

Martijn Pieters

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Comments

Rob Wagner · Accepted Answer · 2013-07-15 18:27:51Z

2

if cont != "No" or cont != "Yes":

Both Yes and No sastisfy this condition

It should be cont != "No" and cont != "Yes"

answered Jul 15, 2013 at 18:27

Rob Wagner

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Comments

llb · Accepted Answer · 2013-07-15 18:28:12Z

2

Cont will always either not equal "No" or not equal "Yes". You want and instead of or.

Or, alternately,

if cont not in ['No', 'Yes']:

which will be more extensible if you want to, e.g., add lowercase.

answered Jul 15, 2013 at 18:28

llb

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Comments

BrenBarn · Accepted Answer · 2013-07-15 18:28:26Z

2

if cont != "No" or cont != "Yes" means "if the answer is not No or it is not Yes". Everything is either not No or not Yes, since it can't be both.

Change it instead to if cont not in ("Yes", "No").

answered Jul 15, 2013 at 18:28

BrenBarn

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Comments

Rohit Jain · Accepted Answer · 2013-07-15 18:28:28Z

2

You need an and operation instead of or. Using or operation, your condition will evaluate to true no matter what the input value is. Change the condition to:

if cont != "No" and cont != "Yes":

or simply use:

if cont not in ("No", "Yes"):

answered Jul 15, 2013 at 18:28

Rohit Jain

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Comments

Jon Kiparsky · Accepted Answer · 2013-07-15 18:30:21Z

-1

if cont != "No" or cont != "Yes":

What could you possibly input that would not satisfy this?

answered Jul 15, 2013 at 18:30

Jon Kiparsky

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