jQuery - keyDown

Question

我遇到了问题，我不明白为什么。任何帮助将不胜感激。

我正在制作我的小引擎，也许是为了一些小游戏。我目前已经为2名玩家设置了它，但我希望将来可以添加更多玩家。我遇到的问题是控件（箭头）仅适用于播放器1.我有播放器2的WSAD，但这不起作用。我已经尝试过调试它，改变它等等。并且无法弄清楚问题。

以下是密钥记录代码：

//========== KEY LOGGING ==========
var pressedKeys = [];
//declare as globals coz of debug
var x;
var y;
var x2;
var y2;

function moveLeft(checkId, checkX, checkY, cSize, cSpeed, cKey) {
  if (x > 0) {
    playerList[checkId].x = checkX - cSpeed;
  } else {
    playerList[checkId].x = 0;
  }
}
function moveUp(checkId, checkX, checkY, cSize, cSpeed, cKey) {
  if (y > 0) {
    playerList[checkId].y = checkY - cSpeed;
  } else {
    playerList[checkId].y = 0;
  }
}
function moveRight(checkId, checkX, checkY, cSize, cSpeed, cKey) {
  if (x2 < width) {
    playerList[checkId].x = checkX + cSpeed;
  } else {
    playerList[checkId].x = width - cSize;
  }
}
function moveDown(checkId, checkX, checkY, cSize, cSpeed, cKey) {
  if (y2 < height) {
    playerList[checkId].y = checkY + cSpeed;
  } else {
    playerList[checkId].y = height - cSize;
  }
}
function Move(checkId, checkX, checkY, cSize, cSpeed, cKey) {
  x = checkX - cSpeed;
  y = checkY - cSpeed;
  x2 = checkX + cSize + cSpeed;
  y2 = checkY + cSize + cSpeed;
  //player 1
  if(checkId == 0) {
    switch (cKey) {
      case 37:
        // left
        moveLeft(checkId, checkX, checkY, cSize, cSpeed, cKey);
        break;
      case 38:
        // up 
        moveUp(checkId, checkX, checkY, cSize, cSpeed, cKey);
        break;
      case 39:
        // right
        moveRight(checkId, checkX, checkY, cSize, cSpeed, cKey);
        break;
      case 40:
        // down
        moveDown(checkId, checkX, checkY, cSize, cSpeed, cKey);
        break;
      default:
        return; // exit this handler for other keys
    }
  }
  //player 2
  if(checkId == 1) {
    switch (cKey) {
      case 65:
        // left - A
        moveLeft(checkId, checkX, checkY, cSize, cSpeed, cKey);
        break;
      case 87:
        // up - W
        moveUp(checkId, checkX, checkY, cSize, cSpeed, cKey);
        break;
      case 68:
        // right - D
        moveRight(checkId, checkX, checkY, cSize, cSpeed, cKey);
        break;
      case 83:
        // down - S
        moveDown(checkId, checkX, checkY, cSize, cSpeed, cKey);
        break;
      default:
        return; // exit this handler for other keys
    }
  }
}

// == KEYDOWN ==
$(document.body).keydown(function (e) {
  e.preventDefault();
  //go through all players
  $.each(playerList, function (i, currentPlayer) {
    if (!pressedKeys[e.which]){
      //set interval for the function
      pressedKeys[e.which] = setInterval(function(){
        Move(currentPlayer.id, currentPlayer.x, currentPlayer.y, currentPlayer.size, currentPlayer.speed, e.which)
      }, 0);
    }
  });
  //+
  if (pressedKeys[107]) {
    currentPlayer.speed += 1; }
    // -
    if (pressedKeys[109] && currentPlayer.speed > 1) {
      currentPlayer.speed -= 1;
    }
    //addplayer
    if (pressedKeys[80]) {
      addPlayer("red", size, width / 2 + id * size, height / 2 + id * size);
    }
  });
  // == KEYUP ==
  $(document.body).keyup(function (e) {
    if (pressedKeys[e.which]){
      clearInterval(pressedKeys[e.which]);
      delete pressedKeys[e.which];
    }
  });

Answer 1

您的具体问题的答案在于：

$.each(playerList, function (i, currentPlayer) {            
  if (!pressedKeys[e.which]){ 
    //set interval for the function
    pressedKeys[e.which] = setInterval(function() {
      Move(currentPlayer.id, currentPlayer.x, currentPlayer.y, currentPlayer.size, currentPlayer.speed, e.which);
    }, 0);
  }
});

让我们说w被推了。此代码遍历播放器列表，从播放器1开始（currentPlayer.id为0）。假设最初pressedKeys为空。由于w被按下，pressedKeys[87]被设置为指向此新间隔的指针，该间隔每0毫秒运行Move。

所以Move运行。但是，您已在Move：

中进行了此项检查

if(checkId == 0) ...

由于'w'仅对玩家2有效（checkId为1），所以没有任何反应，Move会返回。

然后我们回到你的$.each。现在我们开始使用Player 2.然而，我们达到了这个目标：

if (!pressedKeys[e.which]) ...

但pressedKeys[87]已设定。什么都没发生，但它确定了。所以程序会跳过这一步并继续前进。因此，玩家2的任何动作都不会起作用。

你可以这样做：

添加一个数组，其中包含对每个玩家有效的键。在执行if (!pressedKeys[e.which])之前，请检查已按下的键是否有效：

if (validkeys[currentPlayer.id].indexOf(e.which) == -1) return true;