Simple engineering problem, let’s simply it a bit further.

In [1]: a,b,c,d=1,2,3,4
In [2]: a==b
Out[2]: False

But, inherited from the language C, Python evaluates the logical value of a non zero integer as True.

In [11]: if 3:
    ...:     print ("yey")
    ...:
yey

Now, Python builds on that logic and let you use logic literals such as or on integers, and so

In [9]: False or 3
Out[9]: 3

Finally

In [4]: a==b or c or d
Out[4]: 3

The proper way to write it would be:

In [13]: if a in (b,c,d):
    ...:     print('Access granted')

For safety I’d also suggest you don’t hard code passwords.

There are 3 condition checks in if name == "Kevin" or "Jon" or "Inbar":

• name == "Kevin"
• "Jon"
• "Inbar"

and this if statement is equivalent to

if name == "Kevin":
    print("Access granted.")
elif "Jon":
    print("Access granted.")
elif "Inbar":
    print("Access granted.")
else:
    print("Access denied.")

Since elif "Jon" will always be true so access to any user is granted

Solution

---

You can use any one method below

Fast

if name in ["Kevin", "Jon", "Inbar"]:
    print("Access granted.")
else:
    print("Access denied.")

Slow

if name == "Kevin" or name == "Jon" or name == "Inbar":
    print("Access granted.")
else:
    print("Access denied.")

Slow + Unnecessary code

if name == "Kevin":
    print("Access granted.")
elif name == "Jon":
    print("Access granted.")
elif name == "Inbar":
    print("Access granted.")
else:
    print("Access denied.")

Not-empty lists, sets, strings, etc. are evaluable and, therefore, return True.

Therefore, when you say:

a = "Raul"
if a == "Kevin" or "John" or "Inbar":
    pass

You are actually saying:

if "Raul" == "Kevin" or "John" != "" or "Inbar" != "":
    pass

Since at least one of "John" and "Inbar" is not an empty string, the whole expression always returns True!

The solution:

a = "Raul"
if a == "Kevin" or a == "John" or a == "Inbar":
    pass

or:

a = "Raul"
if a in {"Kevin", "John", "Inbar"}:
    pass

Approaches

How a data scientist approaches this problem

The simplest way possible is to eliminate the need for comparison operators and use a list. This looks impressive on security systems because you learn to access ORMs.

user = input("Enter name: ")

if user in {"Bob", "Kevin", "Joe"}:
   print("Access granted, " + str(user) + ".")
else:
   print("Access denied.")

Or, you can resemble the exact same code above, just put the list of registered users in their own list:

user = input("Enter name: ")
users = {"Bob", "Kevin", "Joe", "a million more users if you like"}

if user in users:
   print("Access granted, " + str(user) + ".")
else:
   print("Access denied.")

If you wanted to complete this protocol safely without the risk of attack, set up double parameters. This would check your mini-ORM for first and last name fields, as well as a password or secret question key. Objects can be sorted like this if you want to efficiently lazy-load user credentials without hashing:

def lazy(i):
   j = 0 # For example
   while j < i:
      yield j
      j += 1

The loop will consume only the yielded values to save time and energy on your system:

You can then do something with the iterated list:

for j in lazy_range(10):
   do_something_here(j)

This problem can be approached from any angle: memory management, security, or simply by an organic list or packaged ORM.

Summarising all existing answers

(And adding a few of my points)

Explanation :

if name == "Kevin" or "Jon" or "Inbar":

is logically equivalent to:

if (name == "Kevin") or ("Jon") or ("Inbar"):

Which, for user Bob, is equivalent to:

if (False) or ("Jon") or ("Inbar"):

NOTE : Python evaluates the logical value of any non-zero integer as True. Therefore, all Non-empty lists, sets, strings, etc. are evaluable and return True

The or operator chooses the first argument with a positive truth value.

Therefore, "Jon" has a positive truth value and the if block executes, since it is now equivalent to

if (False) or (True) or (True):

That is what causes "Access granted" to be printed regardless of the name input.

Solutions :

Solution 1 : Use multiple == operators to explicitly check against each value

if name == "Kevin" or name == "Jon" or name == "Inbar":
    print("Access granted.")
else:
    print("Access denied.")

Solution 2 : Compose a collection of valid values (a set, a list or a tuple for example), and use the in operator to test for membership (faster, preferred method)

if name in {"Kevin", "Jon", "Inbar"}:
    print("Access granted.")
else:
    print("Access denied.")

OR

if name in ["Kevin", "Jon", "Inbar"]:
    print("Access granted.")
else:
    print("Access denied.")

Solution 3 : Use the basic (and not very efficient) if-elif-else structure

if name == "Kevin":
    print("Access granted.")
elif name == "Jon":
    print("Access granted.")
elif name == "Inbar":
    print("Access granted.")
else:
    print("Access denied.")

Besides some other rather rarer useful cases for the walrus operator already mentioned. This also tend to be a useful case as well.

def calc_value():
    return 43

if (v := calc_value()) == 43 and v > 42:
    print('happy short, efficient and readable code')

This works because each part of the if-statement is read separately. So (v := calc_value()) is executed and a value is assigned to v and if the first fails, you still have v in the namespace for different conditions or calculations.

Using match/case in Python 3.10 and above

Python 3.10 adds a new syntax to the language. It’s officially described as "structural pattern matching", but most people call it according to the syntax: "match/case".

• Technical specification
• Motivation and rationale (i.e., why it was added, and what inspired the design)
• Official tutorial

We can use this special syntax for an example like in the question, by making one "case" that matches all the accepted usernames, and using the "wildcard" case _ in place of the else. Thus:

name = input("Hello. Please enter your name: ")
match name:
    case "Kevin" | "Jon" | "Inbar":
        print("Access granted.")
    case _:
        print("Access denied.")

Note that cases are "combined" using |, not or. This is a special syntax: Python does not try to compute "Kevin" | "Jon" | "Inbar" first (| doesn’t work with strings), but instead interprets the entire line differently because it starts with case.