在JavaScript中的岩石,纸,剪刀

我正在制作我的第一个游戏(Rock Paper Sissors),我遇到了一个问题,当userChoice是剪刀而且computerChoice是摇滚时,程序无法将胜利者视为摇滚.我可以让程序给我任何其他组合的赢家.

我的代码在这里：

var userChoice = prompt("Do you choose rock,paper or scissors?");
var computerChoice = Math.random();
if (computerChoice < 0.34) {
    computerChoice = "rock";
} else if(computerChoice <= 0.67) {
    computerChoice = "paper";
} else {
    computerChoice = "scissors";
}

var compare = function(choice1,choice2) {
    if(choice1 === choice2) {
    return "The result is a tie!";
}
if(choice1 === "rock") {
    if(choice2 === "scissors") {
        return "rock wins";
    } else {
        return "paper wins";
    }
}
if(choice1 === "paper") {
    if(choice2 === "rock") {
        return "paper wins";
    } else {
        if(choice2 === "scissors") {
            return "scissors wins";
    }
}
if(choice1 === "scissors") {
    if(choice2 === "rock") {
        return "rock wins";
    } else {
        if(choice2 === "paper") {
            return "scissors wins";
        }
    }
}
}
};
console.log("User Choice: " + userChoice);
console.log("Computer Choice: " + computerChoice);
compare(userChoice,computerChoice);

解决方法

由于代码缩进不当,您无法看到问题.正确缩进问题很清楚：

if (choice1 === "paper") {
    if (choice2 === "rock") {
        return "paper wins";
    } else {
        if (choice2 === "scissors") {
            return "scissors wins";
        }
    }
    if (choice1 === "scissors") {
        if (choice2 === "rock") {
            return "rock wins";
        } else {
            if (choice2 === "paper") {
                return "scissors wins";
            }
        }
    }
}

你的if(choice1 ===“剪刀”){在if(choice1 ===“paper”){.永远不会达到内部代码.