javascript - 调试石头剪刀布游戏？ JavaScript

Question

有人可以帮我调试这个简单的石头剪刀布游戏吗？

我认为该错误与 playgame() 函数有关

function playgame() {
  let userChoice = prompt("Rock, Paper, or Scissors?").toUpperCase();
  let computerChoice = computerPlay();
  console.log(rpsgame(computerPlay, userChoice));
}

function computerPlay() {
  return Math.floor(Math.random() * (3) + 1);
  if (computerPlay == 1) {
    computerPlay = "ROCK";
  } else if (computerPlay == 2) {
    computerPlay = "PAPER";
  } else {
    computerPlay = "SCISSORS";
  }
}

Answer 1

return 立即终止函数；您的 computerPlay 函数永远不会超过第一行。相反，您应该将random 调用放入变量中，然后返回ROCK、PAPER 或剪刀。您也不应该尝试将函数名称 (computerPlay) 重新分配给字符串。

function playgame() {
  let userChoice = prompt("Rock, Paper, or Scissors?").toUpperCase();
  let computerChoice = computerPlay();
  console.log(computerChoice);
}

function computerPlay() {
  const rnd = Math.floor(Math.random() * (3));
  if (rnd === 0) {
    return "ROCK";
  } else if (rnd === 1) {
    return "PAPER";
  } else {
    return "SCISSORS";
  }
}
playgame();

如果您想要两者随机数字索引和playgame中的选项名称，您可以立即返回所选的随机数computerPlay 然后让 playgame 将其转换为名称。为了更简洁，使用对象查找而不是 if/else:

function playgame() {
  const playNames = {
    0: 'ROCK',
    1: 'PAPER',
    2: 'SCISSORS'
  }
  let userChoice = prompt("Rock, Paper, or Scissors?").toUpperCase();
  let computerChoice = computerPlay();
  console.log(computerChoice + ' aka ' + playNames[computerChoice]);
}

function computerPlay() {
  return Math.floor(Math.random() * 3);
}
playgame();