python - Always true when testing if string == various OR'ed alternatives

Question

So the problem I am currently having is that my program always calls the 'md5cypher' class which I have defined, even if the input is not in that list:

def enc():
    global toe
    if toe=='md5' or 'M' or 'm' or 'Md5' or 'MD5':
        print("Md5 Encryption Cypher")
        md5cypher()
    else:
        print("Sha1 Encryption Cypher")
        shacypher()

What am I doing wrong?

Answer 1

In reality you're checking:

if (toe=='md5') or 'M' or 'm' or....

And since bool('M') is True, you will always succeed that check. Try this instead:

if toe.lower() in ('md5', 'm'):

Answer 2

if toe=='md5' or 'M' or 'm' or 'Md5' or 'MD5':

will always be evaluated to

if toe=='md5' or True or True or True or True :

What you want is:

if toe in ('md5', 'M', 'm', 'Md5', 'MD5'):
    print("Md5 Encryption Cypher")
    md5cypher()
.
.
.

Answer 3

Literally, what you want to do is:

if toe=='md5' or toe=='M' or toe=='m' or toe=='Md5' or toe=='MD5'

- each part of the composite condition should be a standalone condition.

But in Python you can do it in a more elegant way, as indicated in other answers:

if toe in ('md5', 'M', 'm', 'Md5', 'MD5')

Answer 4

That is because checking a character will always return True therefore ensuring the if statement will execute.

Answer 5

You have a fundamental misunderstanding of how boolean operators work in python.

You have to consider the operator precedences, and then your condition becomes

if (toe=='md5') or ('M') or ('m') or ('Md5') or ('MD5'):

which is equivalent to

if (toe=='md5') or True:

which of course is always true. A solution to your problem would be

if toe.lower() in ('m', 'md5'):