How to test if var is not either of two strings?-CodePudding

I've stumbled upon a problem I can't explain.

chosen = input()
if chosen == "1" or chosen == "2":
  print("Okay")
else:
  print("Please choose between 1 or 2.")

If written like that it executes as intended, but the flow felt weird, so I want to continue with else, so I changed the statement to !=

chosen = input()
if chosen != "1" or chosen != "2":
  print("Please choose between 1 or 2.")
else:
  print("Okay")

That way (to me) it feels natural to continue the code, but now no input returns "Okay".

CodePudding user response：

Ideally, you'd use in for this, which reads much cleaner

while True:
    chosen = input()
    if chosen in ["1", "2"]:
      print("Okay")
      break
    else:
      print("Please choose between 1 or 2.")

CodePudding user response：

You need to use the and keyword instead of or,

by using or this makes it so that if you answer 1 then it will not print "Okay" due to the fact that it is not 2 which makes the function true, and vice versa for the input equaling 2.

This should work:

chosen = input()
if chosen != "1" and chosen != "2":
  print("Please choose between 1 or 2.")
else:
  print("Okay")

Note you may benefit from changing the numbers into integers by doing this:

chosen = int(input())
if chosen != 1 and chosen != 2:
  print("Please choose between 1 or 2.")
else:
  print("Okay")

The int function makes your input into a number, however it will give an error if the user doesn't put in a number.