The program ignores "else" statement

Question

At the beginning it is worth noting that I am a beginner with Python. I've got a problem with my if statement. I'm trying to write a line which will be checking if the player chooses X or O instead of another character. So I used that line to do that:

if player_x_or_o == 'X' or 'O':
    setplayermove(row_choice, col_choice)
else:
    print('Invalid value, try again.')

But if someone uses, for example 'S' the program will do first if instead of else. I can not understand why, any thoughts?

---

# functions


def game_board(X_or_O=0, row=0, column=0, show_gb=False):
    print('     1  2  3')
    if not show_gb:
        game[row][column] = X_or_O
    for index, listof0 in enumerate(game):
        index += 1
        print(index,'.',listof0)

def setplayermove(row, col):
    game[row_choice][col_choice] = player_x_or_o


# Game board

game = [[0,0,0],
        [0,0,0],
        [0,0,0]]

game_board(show_gb=False)

# User input

while True:
    player_x_or_o = str(input('Do you want put "X" or "O"? \n'))
    row_choice = int(input('In which row do you want set your {}? \n'.format(player_x_or_o)))
    col_choice = int(input('In which column do you want set your {}? \n'.format(player_x_or_o)))
    row_choice -= 1
    col_choice -= 1

    if player_x_or_o == 'X' or 'O':
        setplayermove(row_choice, col_choice)
    else:
        print('Invalid value, try again.')



    game_board(show_gb=True)

Answer 1

I think this line is wrong:

if player_x_or_o == 'X' or 'O':

Probably what you meant to write was:

if player_x_or_o == 'X' or player_x_or_o == 'O':

Answer 2

2nd condition is always true, try changing condition as below:

if player_x_or_o == 'X' or player_x_or_o == 'O'