Question

我有一个有5个孩子mcs的容器mc。

儿童姓名为mc0，mc1 .... mc4。

    cont.getChildByName("mc"+Number(cont.numChildren-1)).x = 
            cont.getChildByName("mc0").x - 20 *1.2;

在此重新定位过程之后..我想将最后一个项目位置设置为0，依此类推。我怎么能这样做？

我的目标是实现循环运动。

喜欢

      [mc0][mc1][mc2]
      [mc2][mc0][mc1]
      [mc1][mc2][mc0]
      [mc0][mc1][mc2]

Answer 1

//Of course, you don't necessarily have to create absolute positions,
//this is a simple example...
var positions:Array = [{x:0,y:0} , {x:20, y:20} etc....];
var children:Array = [mc0 , mc1 ... mcN];

//Provided that positions & children have the same length
private function rotate():void
{
    //remove the last element of the Array
    var lastChild:MovieClip = children.pop();

    //Add it to the beginning of the Array
    children.unshift(lastChild );

     //Assign new positions
    //Here you could tween for smoother effect
    for( var i:int ; i < positions.length ; ++i )
    {
      children[i].x = positions[i].x;
      children[i].y = positions[i].y;
    }
}

Answer 2

让我们介绍一个模拟旋转进展的偏移变量：

var offset:uint = 0;

现在我们必须根据此变量定义每个剪辑的位置。我将介绍两个项目之间距离的间隙常数。

const GAP:uint = 20;
for (var iMc:int=0; iMc < cont.numChildren; iMc++)
{
    mc = cont.getChildByName("mc" + iMc.toString()) as Sprite;
    mc.x = GAP * ((iMc + offset) % cont.numChildren);
}

％运算符（modulo）允许您获取介于0和第二个操作数-1之间的数字

如何重新定位MovieClip的子项？

2 个答案: