javascript - 即使我正确填写了提示()，我的 if 语句怎么会不起作用

Question

所以我正在制作一个石头、剪刀、布的小游戏，你可以在其中与电脑进行对抗。

但是，我做了一个 if 语句来检查您是否填写了“石头、剪刀、布”。如果没有，它应该返回一个 alert() ，表明您没有填写这 3 个中的一个，但是当我填写石头、布或剪刀时，它仍然返回 alert().

功能:

function rockPaperScissors() {
    let computerChoice;
    let loopFunction = false;
    let draw = 0;

    while (loopFunction === false) {
        let playerGuess = prompt('Enter rock, paper or scissors!');
        let computerChoiceCalculator = Math.floor(Math.random() * 10);

        if (playerGuess == '') {
            window.alert('ggggg');
            return;
        } else if (playerGuess != 'rock' || 'paper' || 'scissors') {
            window.alert('Please enter rock, paper or scissors');
        }
         else {
            if (computerChoiceCalculator <= 4) {
                computerChoice = 'rock';
                console.log(computerChoice);
            } else if (computerChoiceCalculator <= 7) {
                computerChoice = 'paper';
                console.log(computerChoice);
            } else {
                computerChoice = 'scissors';
                console.log(computerChoice);
            }

            if (playerGuess === computerChoice) {
                draw += 1;
                window.alert('THIS WAS A DRAW! THE COMPUTER HAD ' + computerChoice + ' AND U HAD ' + playerGuess);
            } else if (playerGuess === 'rock' && computerChoice === 'paper' || playerGuess === 'paper' && computerChoice === 'scissors' || playerGuess === 'scissors' && computerChoice === 'rock') {
                window.alert('The computer won this round! Lets try again! your try was ' + playerGuess + ' and the computer had ' + computerChoice);
            } else {
                window.alert('fuck yeah! u won!');
            }
        }
    }
}

导致问题的代码:

else if (playerGuess != 'rock' || 'paper' || 'scissors') {
   window.alert('Please enter rock, paper or scissors');
}

Answer 1

哦，这是错误的:

else if (playerGuess != 'rock' || 'paper' || 'scissors')

此条件不起作用，因为您无法像这样对相等运算符进行分组。

您必须对每个项目进行比较。正如所写的，这个条件将始终返回 true，因为 'paper' 将始终是一个真值。

你应该这样做:

else if (playerGuess != 'rock' && playerGuess != 'paper' && playerGuess != 'scissors')

或者你可以这样做，这样更漂亮:

else if (!['rock', 'paper', 'scissors'].includes(playerGuess))