JavaScript if語句僅返回true

Question

我正在寫剪刀石頭布游戲。

播放器的console.log顯示所選值。

計算機的console.log顯示所選值。

即使控制台日志顯示它們是否不同，用於檢查值是否匹配的if語句也始終返回true。

我嘗試用幾種不同的方式編寫它，但總是會遇到這個問題。

 // Player Choice Selection var player = function() { const playerSelects = "rock"; return playerSelects; }; console.log(player()); // Computer Choice Selection var computer = function() { const computerSelects = ['rock', 'paper', 'scissors']; let randomSelection = Math.floor(Math.random() * 3); return computerSelects[randomSelection]; }; console.log(computer()); // Game var game = function(player, computer) { if (player == computer) { const draw = "its a draw"; return draw; } }; console.log(game()); 

誰能告訴我我要怎么做？

謝謝

Answer 1

游戲是具有兩個參數的函數：

 var game = function(player, computer){
    if(player == computer){

因此，如果您將其稱為

 game()

...如果不傳遞任何值，則player和computer都將是undefined 。可能這樣做：

 game(computer(), player())

...但是您實際上不應該遮蓋這些變量名...

Answer 2

您沒有為game()函數提供任何內容。

我認為這就是您想要的。

如前所述，由於命名的原因， player == computer也不直觀。 人們不會期望玩家能像計算機一樣。 最好有一些類似playerValue == computerValue東西。

 // Player Choice Selection var player = function() { const playerSelects = "rock"; return playerSelects; }; // Computer Choice Selection var computer = function() { const computerSelects = ['rock', 'paper', 'scissors']; let randomSelection = Math.floor(Math.random() * 3); return computerSelects[randomSelection]; }; // Game var game = function(playerValue, computerValue) { if (playerValue === computerValue) { return "its a draw"; } else { return "It's not a draw"; } }; var playerValue = player(); var computerValue = computer(); console.log(playerValue); console.log(computerValue); console.log(game(playerValue, computerValue)); 

如果您不需要函數表達式，則最好使用常規函數，如下所示：

 // Player Choice Selection function getPlayerChoice() { const playerSelects = "rock"; return playerSelects; }; // Computer Choice Selection function getComputerChoice() { const computerSelects = ['rock', 'paper', 'scissors']; let randomSelection = Math.floor(Math.random() * 3); return computerSelects[randomSelection]; }; // Game function playGame(playerChoice, computerChoice) { if (playerChoice === computerChoice) { return "its a draw"; } else { return "It's not a draw"; } }; var playerChoice = getPlayerChoice(); var computerChoice = getComputerChoice(); console.log(playerChoice); console.log(computerChoice); console.log(playGame(playerChoice, computerChoice)); 

Answer 3

您有幾處錯誤。 其他一些問題已部分解決，請允許我解釋一下：

// Player Choice Selection 
var player = function() {
  const playerSelects = "rock";
  return playerSelects;
};
// Don't log the result here, because it will change in the game instance.

// Computer Choice Selection
var computer = function() {
  const computerSelects = ['rock', 'paper', 'scissors'];
  let randomSelection = Math.floor(Math.random() * 3);
  return computerSelects[randomSelection];
};
// Don't log the result here, because it will change in the game instance.

// Game
// Don't pass the player and computer functions in, because they're in global scope already.
var game = function() {
  // Call an instance of the player, and computer functions.
  // We do this so that each game we play, has new results.
  var playerResult = player();
  var computerResult = computer();
  // Log each result
  console.log(playerResult, computerResult)

  if (playerResult == computerResult) {
    const draw = "its a draw";
    return draw;
  }
};
console.log(game());