拖放阵列AS3

Question

我正试图在Actionscript 3中创建一个主要使用数组的拖放游戏。在进入主游戏之前，我会先在一个更简单的游戏中进行，因为我首先需要知道代码是如何工作的。

简单的游戏就是有两个正方形和两个圆形。两个正方形在不同的阵列上，而两个圆在同一阵列中。应该发生的是，当任何一个圆击中(hitTestPoint(右侧的正方形时，它们的x和y就会成为正方形的中心。(就像它点击到中心一样(。当其中一个圆碰到左正方形时，它应该将圆返回到它们的最后位置(不必是它们的原始位置(。

这是代码：

包装｛

import flash.display.MovieClip;
import flash.events.MouseEvent;
import flash.display.DisplayObject;
import flash.geom.Point;
import flash.events.Event;

public class MC_MAIN extends MovieClip
{
var mc1:mc_circle;
var mc2:mc_circle;
var mc3:mc_square;
var mc4:mc_square;
var Shapes:Array;
var Target:Array;
var WTarget:Array;
var newPlace:Point;
public function MC_MAIN()
{
// constructor code
init();
}
function init():void
{
Shapes = new Array  ;
Target = new Array  ;
WTarget = new Array  ;
mc3 = new mc_square();
mc3.height = 75;
mc3.width = 75;
mc3.x = 400;
mc3.y = 200;
Target.push(mc3);
addChild(mc3);
mc4 = new mc_square();
mc4.height = 75;
mc4.width = 75;
mc4.x = 150;
mc4.y = 200;
WTarget.push(mc4);
addChild(mc4);
mc1 = new mc_circle();
mc1.height = 25;
mc1.width = 25;
mc1.x = 100;
mc1.y = 100;
Shapes.push(mc1);
addChild(mc1);
mc2 = new mc_circle();
mc2.height = 25;
mc2.width = 25;
mc2.x = 200;
mc2.y = 200;
Shapes.push(mc2);
addChild(mc2);
for (var i:int = 0; i<Shapes.length; i++)
{
Shapes[i].addEventListener(MouseEvent.MOUSE_DOWN, DRG);
Shapes[i].addEventListener(MouseEvent.MOUSE_UP, SDRG);
}
}
function DRG(e:MouseEvent):void
{
e.currentTarget.startDrag();
}
function SDRG(e:MouseEvent):void
{
e.currentTarget.stopDrag();
for (var m:int = 0; m<Shapes.length; m++)
{
newPlace = new Point(Shapes[m].x,Shapes[m].y);
}
trace(newPlace);
for (var a:int = 0; a<Target.length; a++)
{
for (var b:int = 0; b<Shapes.length; b++)
{
if (Target[a].hitTestPoint(Shapes[b].x,Shapes[b].y))
{
Shapes[b].x = Target[a].x;
Shapes[b].y = Target[a].y;
}
}
}
for (var c:int = 0; c<WTarget.length; c++)
{
for (var d:int = 0; d<Shapes.length; d++)
{
if (WTarget[c].hitTestPoint(Shapes[d].x,Shapes[d].y))
{
Shapes[d].x = newPlace.x;
Shapes[d].y = newPlace.y;
}
}
}
}
}

}

结果是左方块的代码不起作用，但它们没有语法错误。当两个圆圈中的任何一个碰到左边的正方形时，都不会发生任何事情。

当我试图追踪圆的位置时，它只显示了x&y坐标。(我猜它正在跟踪索引为0的数组的第一个对象。我只是想问我是否猜对了这个部分。(

Answer 1

我有点难以遵循逻辑，有些地方没有什么意义，比如：

for (var m:int = 0; m<Shapes.length; m++)
{
newPlace = new Point(Shapes[m].x,Shapes[m].y);
}

newPlace将是Shapes中最后一个shape的位置，因此循环相当无用。

我想你需要的是这样的东西：

public class MC_MAIN extends MovieClip
{
private leftSquares:Array;
private rightSquares:Array;
//more of the members from above
private startPos:Point;
//init the thing and add left and right squares
//to there respective Array
function DRG(e:MouseEvent):void
{
var t:DisplayObject = e.currentTarget;
//save the starting position
startPos = new Point(t.x,t.y);
t.startDrag();
}
function SDRG(e:MouseEvent):void {
var t:DisplayObject = e.currentTarget;
//find all squares from the left
//the target »hits«
var leftHits:Array = leftSquares.filter(
function (square:DisplayObject) {
return square.hitTestPoint(t.x, t.y);
});

//same for the right
var leftHits:Array = rightSquares.filter(
function (square:DisplayObject) {
return square.hitTestPoint(t.x, t.y);
});
//now you can apply the logic
//based on the hit Test result
//this way you can handle the case
//if it hits both, to throw an error
//or alike
if(leftHits.length > 0) {
//reset position
t.x = startPos.x;
t.y = startPos.y;
}
else if (rightHits.length > 0) {
//set the position tp the desired item in rightHits
}
else {
}
}
}

请不要说我的动作脚本技能已经很长时间没有使用了，所以上面的代码可能无法编译。这是为了说明这个想法。重要的是以下步骤：

1. Save the starting position, to be able to reset it
2. Sort the `squares` in respective lists for left and right
3. Hit test both and apply the logic.