conditional statement not running with "or" operator

Question

For context, I'm making a simple number game and asking the user if they would like to play. If the user types "Yes" or "yes" the game should start if the user types "No" or "no" the game will end.

I started with an if/elif/else conditonal but my or operator isn't working so even if the user types "no" it continues the game and runs the next line of code. What am I doing wrong?

Here's my code:

`

if start_game == "Yes" or "yes" :
  print("Let's begin: ")
  print (random_number_one)
  input("Do you think the next number will be higher or lower? Type 'H' for higher or 'L' for lower:  ")
elif start_game == "No" or "no" :
  print("Okay, maybe next time.")
else:
  print("Invalid response. You lose.")

`

I'm not sure why it isn't running correctly and I'm not getting any errors. Is something else wrong?

your syntax is wrong, it should be variable == value1 or variable == value2 — AvirukBasak, Dec 26 '22 at 05:54
Note that the proper way to do a case insensitive comparison is `if start_game.casefold() == "yes":` — MisterMiyagi, Dec 26 '22 at 06:08

score -1 · Accepted Answer · answered Dec 26 '22 at 06:02

Use the following code:

if start_game == "Yes" or start_game == "yes" :
  print("Let's begin: ")
  print (random_number_one)
  input("Do you think the next number will be higher or lower? Type 'H' for higher or 'L' for lower:  ")
elif start_game == "No" or start_game == "no" :
  print("Okay, maybe next time.")
else:
  print("Invalid response. You lose.")

Explanation

Python == operator has precedence over or operator so python sees your code as (start_game == "Yes") or ("yes") and (start_game == "No") or ("no").

A string like "yes" has a truthy value and so is interpreted as True.

So when input is "no", your 1st condition evaluate to (False) or (True) which is True.

conditional statement not running with "or" operator

1 Answers1

Explanation

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