比较字符串值与用户输入的值不正确，

Question

// Computer randomly selects Rock, Paper / Scissors
        computerChoice = function computerPlay() {
            let r = Math.random() * 3;
            let choice;

            if(r > 2) {
                console.log('Rock');
                return 'rock';
            } else if (r > 1) {
                console.log('Scissors');
                return 'scissors';
            } else {
                console.log('Paper');
                return 'paper';
            }
        }


        //compare results
        function compareChoice(userChoice, computerChoice) {
            if (userChoice === computerChoice) {
            console.log('Draw')
            } else if(userChoice === 'paper') {
                if(computerChoice == 'scissors') {
                    console.log ('You lose, try again!')
                } else {
                    console.log('You win!')
                }
            } else if(userChoice === 'rock') {
                if(computerChoice === 'paper') {
                    console.log('You lose! Try again!')
                } else {
                    console.log('You win!')
                }
            } else if(userChoice === 'scissors') {
                if (computerChoice === 'rock') {
                    console.log('You Lose! Try again!')
                } else {
                    console.log('You win!')
                }
            }

        }



        function game() {
            let userChoice = prompt('Choose rock, paper, or scissors!');
            console.log('Your choice was ' + userChoice);
            computerChoice();
            compareChoice();
        }

        game();

你好！我正在尝试为Java中的Odin项目完成一个简单的剪刀石头布项目，但是，每当我尝试将字符串值与用户输入的内容进行比较时，它都会记录日志绘制，这意味着对于第一个if语句，与用户输入的内容无关。我已经搜索过，但不确定自己在做什么错。任何帮助将由衷的感谢！

Answer 1

您需要将用户选择和计算机选择传递到compareChoice中。

    function game() {
        let userChoice = prompt('Choose rock, paper, or scissors!');
        console.log('Your choice was ' + userChoice);
        compareChoice(userChoice, computerChoice());
    }

Answer 2

在您的游戏功能中，您正在不带参数的情况下调用compareChoice()。我不确定该语句如何运行，但唯一的解释似乎是两个参数都被分配了null。因此是“抽奖”响应。

尝试此方法

function game() {
    let userChoice = prompt('Choose rock, paper, or scissors!');
    console.log('Your choice was ' + userChoice);
    compareChoice(userChoice, computerChoice());
}

Answer 3

当您调用compareChoice()函数时，代码不知道userchoice和computerChoice()的值。更改代码，以将这些值作为参数传递给compareChoice()，如下所示：

computerChoice = function computerPlay() {
    let r = Math.random() * 3;
    let choice;

    if (r > 2) {
         console.log('Rock');
         return 'rock';
    } else if (r > 1) {
        console.log('Scissors');
        return 'scissors';
    } else {
        console.log('Paper');
        return 'paper';
    }
}

compareChoice = (userChoice, computerChoice) => {
    if (userChoice === computerChoice) {
        console.log('Draw')
    } else if (userChoice === 'paper') {
        if (computerChoice == 'scissors') {
            console.log ('You lose, try again!')
        } else {
            console.log('You win!')
        }
    } else if (userChoice === 'rock') {
        if(computerChoice === 'paper') {
            console.log('You lose! Try again!')
        } else {
            console.log('You win!')
        }
    } else if (userChoice === 'scissors') {
        if (computerChoice === 'rock') {
            console.log('You Lose! Try again!')
        } else {
            console.log('You win!')
        }
    }
}

game = () => {
    let userChoice = prompt('Choose rock, paper, or scissors!');
    console.log('Your choice was ' + userChoice);
    compareChoice(userChoice, computerChoice());
}

game();