摇滚,纸,剪刀游戏未添加

时间:2019-10-16 02:01:26

标签: javascript

我意识到这是另外一个石头,纸,剪刀的游戏,但是我有点卡住了。我需要这一部分的帮助,然后我想找出其余的部分。

我已经运行了5个循环,现在运行了3个循环。此外,我无法让playerScore和computerScore相加(例如,玩家连续赢得两次,那么playerScore应该等于2)下面是我的代码,我尝试了许多选择,并在互联网上搜寻了几天以试图弄清这一点。如果我有任何语法错误,请告诉我,我们会先解决,然后再发布。谢谢!

var playerScore = 0;
var computerScore = 0;

function computerPlay() {
  const cptchoice = ["Rock", "Paper", "Scissors"];
  const randomNumber = Math.floor(Math.random() * 3);
  return cptchoice[randomNumber];
}

function playerPlay() {
  const playerPrompt = prompt("Paper, Scissors, or Rock");
  return playerPrompt[0].toUpperCase() + playerPrompt.slice(1).toLowerCase();
}

var playRound = function() {
  for (var i = 0; i < 5; i++) {
    const playerSelection = playerPlay();
    const computerSelection = computerPlay();
    switch (playerSelection + computerSelection) {
      case "PaperRock":
      case "RockScissors":
      case "ScissorsPaper":
        console.log("Win");
        return playerScore + 1;
        break;
      case "PaperScissors":
      case "RockPaper":
      case "ScissorsRock":
        console.log("Lose");
        return computerScore + 1;
        break;
      case "PaperPaper":
      case "RockRock":
      case "ScissorsScissors":
        console.log("Tie");
        console.log(playerScore + 0);
    }
    console.log(playerScore);
  }
};

1 个答案:

答案 0 :(得分:0)

您正在switch case语句中调用System.out.println( javax.crypto.Cipher.getMaxAllowedKeyLength("AES"));,这导致函数return完成执行。如果仅删除playRound()并仅添加到乐谱中,它将一直进行下去。

return