在AS3中分配一組對象的寬度

Question

假設我有一組具有相同y位置的5個動畫片段，但按升序分佈x個位置（例如obj1.x = 0,obj5.x = 10），是否有AS3方法可幫助我分配它們的寬度像下修改>對齊閃光燈的選項，所以它們的x位置0和10之間等距間隔？

感謝

Answer 1

這是一個功能我多年前寫的，你可能會發現有用的。它假設你想要放置的東西是父母/容器的唯一孩子。您將該父項作爲第一個參數（displayObj）傳遞給此函數。希望在最開始的註釋向你解釋不夠好各參數做什麼，如果不只是評論，我會澄清。

/** 
* Distribte all the children of the specified display object on either x or y axis 
* @param displayObj - the display oject whose children should be distributed 
* @param onX - should it distribute along the x axis (true) or the y axis (false) 
* @param spacing - how much space between children in pixels 
* @param center - should the children be centered on the opposite axis 
* @param startingX - default 0 
* @param startingY - default 0 
* @param fixedWidth - use a fixed width instead the automatic width of each child 
* @param foldPoint - how far along before the item should fold and be a new row/col 
*/ 
public static function distributeAxis(displayObj:Sprite,onX:Boolean = true,spacing:Number = 5, center:Boolean = false, startingX:Number = 0,startingY:Number = 0,fixedWidth:Number = 0, foldPoint:Number = 0):void { 

    //Loop Through Children 
     var tObj:DisplayObject; 
     var tNum :Number = (onX) ? startingX : startingY; 
     var tNum2 :Number = (onX) ? startingY : startingX; 
     var max  :Number = 0; 

     var centeringArray:Vector.<DisplayObject> 
     if (center) { 
      centeringArray = new Vector.<DisplayObject>(); 
     } 

     for(var i:int = 0; i<displayObj.numChildren;i++) 
     { 
      tObj = displayObj.getChildAt(i); 

      if (onX) { 
       if (foldPoint > 0 && tNum + tObj.width > foldPoint) { 
        tNum = startingX; 
        tNum2 += max + spacing; 

        if(center){ 
         distributeAxisCenterIt(centeringArray, max, onX); 
         centeringArray = new Vector.<DisplayObject>(); 
        } 

        max = 0; 
       } 

       if(tObj.height > max) max = tObj.height; 

       tObj.x = tNum; 
       tObj.y = tNum2; 

       if(fixedWidth > 0){ 
        tNum += fixedWidth + spacing; 
       }else{ 
        tNum += tObj.width + spacing; 
       } 

       if(center){ 
        centeringArray.push(tObj); 
       } 

      }else{ 
       if(tObj.width > max) max = tObj.width; 

       if (foldPoint > 0 && tNum + tObj.height > foldPoint) { 
        tNum = startingY; 
        tNum2 += max + spacing; 

        if(center){ 
         distributeAxisCenterIt(centeringArray, max, onX); 
         centeringArray = new Vector.<DisplayObject>(); 
        } 
        max = 0; 
       } 

       tObj.y = tNum; 
       tObj.x = tNum2; 

       if(fixedWidth > 0){ 
        tNum += fixedWidth + spacing; 
       }else{ 
        tNum += tObj.height + spacing; 
       } 

       if(center){ 
        centeringArray.push(tObj); 
       } 
      } 

     } 


    if (center) { 
     distributeAxisCenterIt(centeringArray, max, onX); 
    } 
} 

private static function distributeAxisCenterIt(array:Vector.<DisplayObject>, max:Number, onX:Boolean = true):void { 
       for each(var tObj:DisplayObject in array){ 
        if(onX){ 
         tObj.y += ((max - tObj.height) * .5); 
        }else{ 
         tObj.x += ((max - tObj.width) * .5); 
        } 

       } 
      } 

Answer 2

你需要使用這個：AS3 Commons UI。它擁有所有你所能想:)與佈局/對齊等和組件

，但如果你想有一個代碼，沒有一個框架，這是我使用的是什麼，我自己的「框架」會在不同的寬度擴展項目：

/** 
    * Distributes all components using given value as a limit. 
    * @param p_nMaxWidth width of the area to spread items 
    */ 
    protected function spreadComponents(p_nMaxWidth:Number):void 
    { 
     if (numChildren == 0) return; 

     if (p_nMaxWidth < 0 || isNaN(p_nMaxWidth)) p_nMaxWidth = 600; 
     var _oItem:DisplayObject; 
     var dx:Number = 0; 
     var i:int; 
     var _nWidth:Number = calculateWidths();//get sum of all items widths 
     var _nStep:Number = (p_nMaxWidth - _nWidth - getChildAt(numChildren - 1).width)/(numChildren-1); 
     for (i = 0; i < numChildren; i++) 
     { 
      _oItem = getChildAt(i); 
      //if(m_bResetChildrenPosition) _oItem.y = 0;//this I was using to reset or keep intact the items y position 
      _oItem.x = Math.round(dx); 
       dx += _oItem.width + _nStep; 
     } 
     //as a precaution if rounding would give incorrect position (which should be for last item exactly at the p_nMaxWidth - width (right edge aligned)). 
     getChildAt(numChildren - 1).x = p_nMaxWidth - getChildAt(numChildren - 1).width; 

     updateDimension(); 
    } 


// 
    /** 
    * Utility method, returns width of all children (sum). 
    * @return 
    */ 
    protected function calculateWidths():Number 
    { 
     var _nWidth:Number = 0; 
      for (var i:int = 0; i < numChildren-1; i++) 
      { 
       _nWidth += getChildAt(i).width; 
      } 
     return _nWidth; 
    } 

這個碼擴展項目horizo​​ntaly會在不同的寬度 - 第一個項目的左側邊緣在啓動，最後一項的右邊緣位於認爲「寬度」結束。

最好的問候