1

我正在创建一个鱼缸..所以我想做的是我的鱼应该是无限滚动的..我正在使用动作脚本 3..这是我使用的代码..但它不起作用..

var scrollSpeed:uint = 5;

//This adds two instances of the movie clip onto the stage.



//This positions the second movieclip next to the first one.
f1.x = 0;
f2.x = f1.width;

//Adds an event listener to the stage.
stage.addEventListener(Event.ENTER_FRAME, moveScroll); 

//This function moves both the images to left. If the first and second 
//images goes pass the left stage boundary then it gets moved to 
//the other side of the stage. 
function moveScroll(e:Event):void{
f1.x -= scrollSpeed;  
f2.x -= scrollSpeed;  

if(f1.x < +f1.width){
f1.x = f1.width;
}else if(f2.x < +f2.width){
f2.x = f2.width;
}
}

f1 和 f2 是我的鱼实例名称

4

2 回答 2

1

你的问题线在底部。

if (f1.x < f1.width) {
    f1.x = stage.stageWidth - f1.width // f1.width;
} // Removed else

if (f2.x < f2.width) {
    f2.x = f1.width;
}
于 2013-07-08T10:50:13.630 回答
0

这假设您的图像/mc 对于屏幕来说足够大，并且是重复的（具有相同的内容）

function moveScroll(e:Event):void {
    //position fish 1
    f1.x -= scrollSpeed;
    //when fish 1 is out of bounds (off the screen on the left, set it back to 0
    if (f1.x < -f1.width) {
       f1.x = 0;
    }
    //always position fish2 next to fish 1
    f2.x = f1.x + f1.width;
}

另一种方法是将它们放在一个数组中并切换它们，如果它们没有相同的内容：

//add fishes to array
var images:Array = new Array();
images.push(f1, f2);

function moveScroll(e:Event):void {
    //position fish 1
    images[0].x -= scrollSpeed;
    //when first fish is out of bounds, remove it from the array (unshift) and add it at the end (push)
    if (images[0].x < -images[0].width) {
       images.push(images.unshift())
    }
    //always position fish2 next to fish 1
    images[1].x = images[0].x + images[0].width;
}

这些只是基本的例子，但你明白了

于 2013-07-08T08:57:16.143 回答