Actionscript 3 AS3.0:从第二类访问已创建实例的属性

我正在制作一个小游戏

在我的document类中，我使用以下代码创建类

字符

和

级别

的实例：

//add the Level
level = new TileGrid();
level.y = 100;
level.x = 400;
addChild(player);

//add our player
player = new Character();
player.y = 150;
player.x = 400;
addChild(player);

我还创建了一个控制器类来处理用户输入。（例如，检查玩家是否能够向右移动。）

我还为键盘事件等创建事件监听器

当按下一个键时，我想通过在控制器类中调用TileGrid类的

checkTile（tileNumber）

函数来检查移动是否可行

控制器类如下所示：

package  {
    import flash.events.KeyboardEvent;
    import flash.events.Event;

    public class Controller{

    //Constructor code
    public function Controller(){}

    //Keyboard pressed -> move character
    public function keyPressed(evt:KeyboardEvent):void
    {
        trace(level.checkTile(30));
    }
}

package  {
    import flash.events.KeyboardEvent;
    import flash.events.Event;

    public class TileGrid{

    //Constructor code
    public function TileGrid(){
        //Creating all the tiles and adding them to the stage.
    }

    //Check if a certain tile is walkable
    public function checkTile(tileNumberType){
        if(tileNumberType > 15){
            return false;
        }else{
           return true;
        }
    }
}

...
// somewhere in your document class (or somewhere else)
var player:Character = new Character();
var level:TileGrid = new TileGrid();
var controller:Controller = new Controller(player, level);
...

// in your Controller class

private var level:TileGrid;
private var player:Character;

public Controller(player:Character, level:TileGrid) {
    this.player = player;
    this.level = level;
}

public function keyPressed(event:KeyboardEvent):void {
    level.checkTile(30); // in this line "level" means "this.level"
}

TileGrid类看起来像这样：

package  {
    import flash.events.KeyboardEvent;
    import flash.events.Event;

    public class Controller{

    //Constructor code
    public function Controller(){}

    //Keyboard pressed -> move character
    public function keyPressed(evt:KeyboardEvent):void
    {
        trace(level.checkTile(30));
    }
}

package  {
    import flash.events.KeyboardEvent;
    import flash.events.Event;

    public class TileGrid{

    //Constructor code
    public function TileGrid(){
        //Creating all the tiles and adding them to the stage.
    }

    //Check if a certain tile is walkable
    public function checkTile(tileNumberType){
        if(tileNumberType > 15){
            return false;
        }else{
           return true;
        }
    }
}

...
// somewhere in your document class (or somewhere else)
var player:Character = new Character();
var level:TileGrid = new TileGrid();
var controller:Controller = new Controller(player, level);
...

// in your Controller class

private var level:TileGrid;
private var player:Character;

public Controller(player:Character, level:TileGrid) {
    this.player = player;
    this.level = level;
}

public function keyPressed(event:KeyboardEvent):void {
    level.checkTile(30); // in this line "level" means "this.level"
}

但是当我测试这个时，我得到以下错误：

第81 1120行：访问未定义的属性级别。

当我尝试时：

trace（Object（parent）.level.checkTile（30））我获取：
1120:访问未定义的属性父项。

如何从一个类访问方法，从另一个类访问实例
 AS中的类无法访问其上下文。这意味着如果我有：

function foo():void
{
    var k:MyCustomClass = new MyCustomClass();
    var j:MyOtherClass  = new MyOtherClass();
}

这两个例子j和k彼此不了解

但是，在这种特殊情况下，因为您显然在处理父结构，所以可以获取父结构的属性。这是可能的，因为您在公共作用域中有一个变量的路径

请尝试以下方法：

trace(Object(parent).level.checkTile(30));

还要注意对象（父对象）。DisplayObject的父对象是DisplayObjectContainer，它没有级别属性。然而，通过在对象中包装父对象，您告诉Flash，“没关系，您应该在运行时而不是在编译时查找此属性。”
我认为您必须这样做：

package  {
    import flash.events.KeyboardEvent;
    import flash.events.Event;

    public class Controller{

    //Constructor code
    public function Controller(){}

    //Keyboard pressed -> move character
    public function keyPressed(evt:KeyboardEvent):void
    {
        trace(level.checkTile(30));
    }
}

package  {
    import flash.events.KeyboardEvent;
    import flash.events.Event;

    public class TileGrid{

    //Constructor code
    public function TileGrid(){
        //Creating all the tiles and adding them to the stage.
    }

    //Check if a certain tile is walkable
    public function checkTile(tileNumberType){
        if(tileNumberType > 15){
            return false;
        }else{
           return true;
        }
    }
}

...
// somewhere in your document class (or somewhere else)
var player:Character = new Character();
var level:TileGrid = new TileGrid();
var controller:Controller = new Controller(player, level);
...

// in your Controller class

private var level:TileGrid;
private var player:Character;

public Controller(player:Character, level:TileGrid) {
    this.player = player;
    this.level = level;
}

public function keyPressed(event:KeyboardEvent):void {
    level.checkTile(30); // in this line "level" means "this.level"
}

在这种情况下，您必须说明控制器必须控制哪个玩家和哪个级别。控制器是一个不了解任何其他类的类。这些变量不是全局变量（在您的示例中，它们不应该是全局变量），因此您无法从任何地方访问它们。
我尝试过，但没有成功，我得到了错误：1120:访问未定义的属性父对象。这将不起作用，因为您可能认为
Controller
是添加到stage的DisplayObject，而事实并非如此。