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我有一种在 Bejeweled 中交换两件作品的方法，但是当用户进行没有匹配的交换时，动画交换不会出现在用户面前。

即当用户尝试交换宝石并且没有匹配发生时，宝石保持静止。应该发生的是宝石交换位置，意识到没有匹配并交换回来。

我认为问题在于，如果没有匹配项并且用户没有动画，则交换几乎立即发生。

如何减慢第二次交换（即交换回）的执行时间，以便发生交换动画？

// start animated swap of two pieces

public function makeSwap(piece1,piece2:Piece)
{
    swapPieces(piece1,piece2);

    //check to see if move works

    if (lookForMatches().length == 0)
    {
        swapPieces(piece1,piece2); //Swap the piece back
    } 
    else 
    {
        isSwapping = true;
    }
}

// swap two pieces

public function swapPieces(piece1,piece2:Piece) 
{
    // swap row and col values

    var tempCol:uint = piece1.col;
    var tempRow:uint = piece1.row;
    piece1.col = piece2.col;
    piece1.row = piece2.row;
    piece2.col = tempCol;
    piece2.row = tempRow;

    // swap grid positions

    grid[piece1.col][piece1.row] = piece1;
    grid[piece2.col][piece2.row] = piece2;

}
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1 回答 1

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从您的代码中并不太清楚动画本身是如何发生的（显然在其他地方有渲染函数），但是要在 AS3 中为函数的执行添加延迟，您可以使用setTimeout函数或Timer类。

这是setTimeout：

import flash.utils.setTimeout; //To be added with your other imports

// start animated swap of two pieces

public function makeSwap(piece1:Piece,piece2:Piece):void
{
    swapPieces(piece1,piece2);

    //check to see if move works

    if (lookForMatches().length == 0)
    {
        setTimeout(swapPieces, 500, piece1, piece2); //Swap the piece back, after delay of 500 milliseconds
    } 
    else 
    {
        isSwapping = true;
    }
}

// swap two pieces

public function swapPieces(piece1:Piece,piece2:Piece):void
{
    // swap row and col values

    var tempCol:uint = piece1.col;
    var tempRow:uint = piece1.row;
    piece1.col = piece2.col;
    piece1.row = piece2.row;
    piece2.col = tempCol;
    piece2.row = tempRow;

    // swap grid positions

    grid[piece1.col][piece1.row] = piece1;
    grid[piece2.col][piece2.row] = piece2;

}

这是Timer：

import flash.utils.Timer; //To be added with your other imports
import flash.events.TimerEvent; //To be added with your other imports

private var _swapBackTimer:Timer = new Timer(500, 1); //Delay of 500 milliseconds, timer running 1 time
private var _delaySwapPiece1:Piece;
private var _delaySwapPiece2:Piece;

// start animated swap of two pieces

public function makeSwap(piece1:Piece,piece2:Piece):void
{
    swapPieces(piece1,piece2);

    //check to see if move works

    if (lookForMatches().length == 0)
    {
        _swapBackTimer.reset();
        _swapBackTimer.addEventListener(TimerEvent.TIMER, swapBackTimer_complete);
        _swapPiece1 = piece1;
        _swapPiece2 = piece2;
        _swapBackTimer.start();
    } 
    else 
    {
        isSwapping = true;
    }
}

private function swapBackTimer_complete(evt:TimerEvent):void
{
    _swapBackTimer.removeEventListener(TimerEvent.TIMER, swapBackTimer_complete);
    swapPieces(_delaySwapPiece1, _delaySwapPiece2);
}

// swap two pieces

public function swapPieces(piece1:Piece,piece2:Piece):void
{
    // swap row and col values

    var tempCol:uint = piece1.col;
    var tempRow:uint = piece1.row;
    piece1.col = piece2.col;
    piece1.row = piece2.row;
    piece2.col = tempCol;
    piece2.row = tempRow;

    // swap grid positions

    grid[piece1.col][piece1.row] = piece1;
    grid[piece2.col][piece2.row] = piece2;

}

需要记住的一件事是用户在延迟期间更改游戏状态、使延迟操作无效和引入错误的风险。您可能已经在管理这个，但以防万一，一种方法是在任何延迟操作期间设置一个标志，然后在此期间忽略任何后续输入。另一种方法是取消延迟的操作，让它立即执行，然后从该点开始处理新的用户操作。

于 2013-09-06T09:19:07.673 回答