0

晚上好，

我似乎无法让我的代码只向舞台添加一个对象实例。它似乎在舞台上添加了至少 3 个对象实例。添加的对象是 r2、r3 和 r4。

stop();

import flash.events.Event;
stage.focus=stage;
var upKeyDown5:Boolean = false;
var rightKeyDown5:Boolean = false;
var downKeyDown5:Boolean = false;
var leftKeyDown5:Boolean = false;
var enterkey:Boolean = false;
var interaction:Boolean = false;
p5.addEventListener(Event.ENTER_FRAME, moveChar5);
stage.addEventListener(KeyboardEvent.KEY_DOWN, checkKeysDown5);
stage.addEventListener(KeyboardEvent.KEY_UP, checkKeysUp5);
Mouse.hide();
function moveChar5(e:Event):void{
if(downKeyDown5 && !upKeyDown5 && !rightKeyDown5 && !leftKeyDown5)
{
    p5.gotoAndStop("walk_down");
    if(p5.y < 492.75)
        p5.y += 6;
}
if(upKeyDown5 && !downKeyDown5 && !rightKeyDown5 && !leftKeyDown5)
{
    p5.gotoAndStop("walk_up");
    if(p5.y > 202.85)
        p5.y -= 6;
}
if(rightKeyDown5 && !upKeyDown5 && !downKeyDown5 && !leftKeyDown5)
{
    p5.gotoAndStop("walk_right");
    if(p5.x < 871.5)
        p5.x += 6;
}
if(leftKeyDown5 && !upKeyDown5 && !rightKeyDown5 && !downKeyDown5)
{
    p5.gotoAndStop("walk_left");
    if(p5.x > 203.65)
        p5.x -= 6;
}
if(enterkey && interaction && p5.hitTestObject(c1)){
    if (!(Boolean(stage.getChildByName('r2')))) {
        var rat2:r2;
        rat2 = new r2();
        addChild(rat2);
        rat2.y=38;
        rat2.x=32;
    }
}
if(enterkey && interaction && p5.hitTestObject(c2)){
    if (!(Boolean(stage.getChildByName('r3')))) {
        var rat3:r3;
        rat3 = new r3();
        addChild(rat3);
        rat3.y=38;
        rat3.x=32;
    }
}
if(enterkey && interaction && p5.hitTestObject(c3)){
    if (!(Boolean(stage.getChildByName('r4')))) {
        var rat4:r4;
        rat4 = new r4();
        addChild(rat4);
        rat4.y=38;
        rat4.x=32;
    }
}
 }


 function checkKeysDown5(event:KeyboardEvent):void{
if(event.keyCode == 87){
    upKeyDown5 = true;
}
if(event.keyCode == 68){
    rightKeyDown5 = true;
}
if(event.keyCode == 83){
    downKeyDown5 = true;
}
if(event.keyCode == 65){
    leftKeyDown5 = true;
}
if (event.keyCode == 13){
    enterkey = true;
}
 }


 function checkKeysUp5(event:KeyboardEvent):void{
if(event.keyCode == 87){
    upKeyDown5 = false;
    p5.gotoAndStop("still_up");
}

if(event.keyCode == 68){
    rightKeyDown5 = false;
    p5.gotoAndStop("still_right");
}

if(event.keyCode == 65){
    leftKeyDown5 = false;
    p5.gotoAndStop("still_left");
}

if(event.keyCode == 83){
    downKeyDown5 = false;
    p5.gotoAndStop("still_down");
}

if (event.keyCode == 13){
    enterkey = false;
}

if(p5.hitTestObject(c1) || p5.hitTestObject(c2) || p5.hitTestObject(c3)){
    p5.gotoAndStop("interaction");
    interaction = true;
}
else
    interaction = false;
 }

提前致谢

4

2 回答 2

2

使用语句而不是所有那些 if 语句会更有效和更switch简洁（如果您只希望满足其中一个且仅满足其中一个条件）

我认为您的问题与您跟踪老鼠的方式有关-例如。使用stage.getChildByName.
下面是一个示例，以及一些关于您的其他一些潜在问题的注释。

switch(true){
    case (enterkey && interaction && p5.hitTestObject(c1)):
        if (!(Boolean(stage.getChildByName('r2')))) {  //your not giving your rat a name so this will always return false, plus you're not adding the rat to the stage but to this class
            var rat2:r2;
            rat2 = new r2();
            addChild(rat2); //use stage.addChild if you want the above check (stage.getChildByName) to work
            rat2.y=38;
            rat2.x=32;
            rat2.name = "r2"; //if you want the above check to work

            break;  //break out of the switch if you don't want any of the others evaluate since this rat got added
        }

    //do a case for your other blocks
}

最好不要使用 stage.getChidByName ，而是创建一个这样的方法：

function checkRatExists(ratClass:Class):Boolean {
    var tmp:DisplayObject;
    var i:int = numChildren;
    while(i--){
        tmp = getChildAt(i);
        if(tmp is ratClass){
            return true;
        }
    }
    return false;
}

然后使用该新函数而不是 stage.getChildByName：

if (!checkRatExists(r2)) //r2 is the class of rat you want to check

除了你的问题之外，创建一个方法/函数来做你的老鼠定位和添加而不是一遍又一遍地重复代码会更干净......

function addRat(rat:RatCommonBaseClass, ratName:String):void {
        addChild(rat);
        rat.y=38;
        rat.x=32;
        rat.name = ratName;
}

//now you can just do this for all your rats:
addRat(new r2(),"r2);
于 2012-11-01T16:42:39.307 回答
0

对于添加对象的语句，您没有 else 条件，因此很可能发生的情况是所有语句都同时满足条件。添加 else 语句应该可以防止这种情况发生：

if(enterkey && interaction && p5.hitTestObject(c1)){
    if (!(Boolean(stage.getChildByName('r2')))) {
        var rat2:r2;
        rat2 = new r2();
        addChild(rat2);
        rat2.y=38;
        rat2.x=32;
    }
}
else if(enterkey && interaction && p5.hitTestObject(c2)){
    if (!(Boolean(stage.getChildByName('r3')))) {
        var rat3:r3;
        rat3 = new r3();
        addChild(rat3);
        rat3.y=38;
        rat3.x=32;
    }
}
else if(enterkey && interaction && p5.hitTestObject(c3)){
    if (!(Boolean(stage.getChildByName('r4')))) {
        var rat4:r4;
        rat4 = new r4();
        addChild(rat4);
        rat4.y=38;
        rat4.x=32;
    }
}

这样，它将首先检查 p5 是否已击中 c1，只有在没有击中时才会检查是否已击中 c2，然后对 c3 进行相同的操作。

于 2012-11-01T16:43:27.403 回答