Why writing a correct input uppercase throws an error?

Question

If user inputs a correct word uppercase I get a "You made a mistake" response. Why does user_choice.lower() not work like I intended? Where did I make a mistake?

import random

rock_paper_scissors = ("rock", "paper","scissors")
cpu_choice = random.choice(rock_paper_scissors)
user_choice = input("Rock, paper or scissors ? Enter your choice: ")


if user_choice.lower() != "rock" or "paper" or "scissors":
    print("You made a mistake. Please try again")
    user_choice = input("Rock, paper or scissors ? Enter your choice: ")

while cpu_choice == user_choice.lower():
    print ("--------------------------------------------------------")
    print(f"CPU: {cpu_choice.upper()} vs YOU: {user_choice.upper()}.  It's a tie. Try again!")
    print ("--------------------------------------------------------")
    user_choice = input("Rock, paper or scissors ? Enter your choice: ")

`if user_choice.lower() != "rock" or "paper" or "scissors"` is not the right way to check for more than one value. See this question https://stackoverflow.com/q/15112125/494134 — John Gordon, Dec 14 '20 at 15:05

quamrana · Answer 1 · 2020-12-14T15:19:24.250

2

Did you mean:

if user_choice.lower() not in rock_paper_scissors:
    print("You made a mistake. Please try again")

edited Dec 14 '20 at 15:19

answered Dec 14 '20 at 15:08

quamrana

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Maybe even better: `if user_choice.lower() not in rock_paper_scissors:`, so we don't have to write those values again. With just three values performance is not an issue. – Matthias Dec 14 '20 at 15:15

score 1 · Accepted Answer · answered Dec 14 '20 at 15:06

1

The condition if you set isn't correct. Try instead

if user_choice.lower() != "rock" or user_choice.lower() != "paper" or user_choice.lower() != "scissors":

answered Dec 14 '20 at 15:06

mhhabib

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Why writing a correct input uppercase throws an error?

2 Answers2