javascript - 同时监听多个按键事件

Question

好吧，所以我正在开发 Agar 的简单克隆但只有2名玩家。我已经掌握了大部分基本功能，但我不知道如何同时处理两个玩家的按键事件。 (一次只能移动一个)这是我的 jsFiddle 。相关片段:

if (keysArr.toString() === "37,87") {
    player.dir = "left";
    player2.dir = "up";
    keys = [];
} 

if (keysArr.toString() === "38,87") {
    player.dir = "up";
    player2.dir = "up";
}

if (keysArr.toString() === "39,87") {
    player.dir = "right";
    player2.dir = "up";
}

if (keysArr.toString() === "40,87") {
    player.dir = "down";
    player2.dir = "up";
} 

if (keysArr.toString() === "37,65") {
    player.dir = "left";
    player2.dir = "left";
} 

if (keysArr.toString() === "38,65") {
    player.dir = "up";
    player2.dir = "left";
} 

if (keysArr.toString() === "39,65") {
    player.dir = "right";
    player2.dir = "left";
} 

if (keysArr.toString() === "40,65") {
    player.dir = "down";
    player2.dir = "left";
} 

if (keysArr.toString() === "37,83") {
    player.dir = "left";
    player2.dir = "down";
} 

if (keysArr.toString() === "38,83") {
    player.dir = "up";
    player2.dir = "down";
} 

if (keysArr.toString() === "39,83") {
    player.dir = "right";
    player2.dir = "down";
} 

if (keysArr.toString() === "40,83") {
    player.dir = "down";
    player2.dir = "down";
} 

if (keysArr.toString() === "37,68") {
    player.dir = "left";
    player2.dir = "right";
} 

if (keysArr.toString() === "38,68") {
    player.dir = "up";
    player2.dir = "right";
} 

if (keysArr.toString() === "39,68") {
    player.dir = "right";
    player2.dir = "right";
} 

if (keysArr.toString() === "40,68") {
    player.dir = "down";
    player2.dir = "right";
} 

if (keysArr.toString() === "37") {
    player.dir = "left";
} 

if (keysArr.toString() === "38") {
    player.dir = "up";
} 

if (keysArr.toString() === "39") {
    player.dir = "right";
} 

if (keysArr.toString() === "40") {
    player.dir = "down";
}

我想不出其他方法，所以我只是使用了一堆 if。

所以，这段代码可以让两个玩家同时移动，但在第一个玩家之后，他们就不能再转动了。我该如何做才能让两者同时移动并随后转动？

Answer 1

正在检测按住第一个键时按下的第二个键，因为其他玩家将移动。

您必须将所有 ELSEIF 更改为简单的 IF。将玩家 2 elseif 与玩家 1 if 绑定(bind)意味着如果玩家 1 按下某个键，它将永远不会达到玩家 2 的 elseif 条件。这将解决您的主要问题。

document.onkeydown 和 document.onkeyup 监听器不需要每次重复都初始化，只需初始化一次。

我修改了您的 jsFiddle 以反射(reflect)这些更改。然后我得意忘形，还将代码放在 document.onkeyup 中，以使玩家停止移动，因为这让我很困扰。享受! https://jsfiddle.net/wtqk7Lu2/1/

if (player2.dir === keys.w) {
    player2.y -= player2.speed.y;
}
if (player.dir === keys.up) {
    player.y -= player.speed.y;
}
if (player2.dir === keys.a) {
    player2X -= player2.speed.x;
}
if (player.dir === keys.left) {
    player1X -= player.speed.x;
}
if (player2.dir === keys.s) {
    player2.y += player2.speed.y;
}
if (player.dir === keys.down) {
    player.y += player.speed.y;
}
if (player2.dir === keys.d) {
    player2X += player2.speed.x;
}
if (player.dir === keys.right) {
    player1X += player.speed.x;
}