jQuery - keyDown

Question

我有一个问题，我不知道为什么。任何帮助都将不胜感激。

我正在做我的小引擎，也许是为了一些小游戏。我目前有它为2名球员设置，但我希望能够添加更多在未来。我的问题是，控件(箭头)只对玩家1有效。我有player2的WSAD，但它不起作用。我试着调试它，改变它，等等。找不到问题所在。

以下是键日志记录代码：

//========== 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

你的具体问题的答案在这里：

$.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);
  }
});

假设w被推送。这段代码迭代球员列表，从球员1开始(currentPlayer.id为0)。假设最初pressedKeys是空的。由于按下了w，因此pressedKeys[87]被设置为指向这个新时间间隔的指针，它每隔0毫秒运行一次Move。

于是Move开始运行。但是，您可以在Move中执行此检查

if(checkId == 0) ...

因为'w‘只对播放器2有效(checkId为1)，所以不会发生任何事情，Move会返回。

然后我们回到你的$.each。现在我们可以和玩家2一起工作了。但是，我们可以这样做：

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

但是pressedKeys[87]已经设置好了。什么都没发生，但一切都安排好了。因此，程序跳过这一步，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：

添加一个包含对每个玩家有效的密钥的数组。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;