有几种方法。这取决于你有什么,可以和不能在那里使用哪一个。
方式№1。您可以为所有字母使用单独的容器并准确清理该容器。
// Before your code.
var TheABC:Sprite = new Sprite;
addChild(TheABC);
// Inside the letter creation methods.
// I do hope you do not have 26 separate functions for that.
// ...
TheABC.addChild(LetraA_1);
// ...
// ...
TheABC.addChild(LetraB_1);
// ...
// So, the finale is very simple.
function eliminarhijos(e:MouseEvent):void
{
// Remove all the children at once.
TheABC.removeChildren();
// If, by any chance, you have a VERY old Flash
// below Player 11, the one above won't work so
// you would have to resolve to the one below.
// Proceed while there are children at all.
// while (ThABC.numChildren)
// {
// // Remove the firstmost child.
// TheABC.removeChildAt(0);
// }
}
方式№2。报名名单。您保留一份需要删除的内容的列表。
// Before your code.
var ABClist:Array = new Array;
// Inside the letter creation methods.
// ...
addChild(LetraA_1);
ABClist.push(LetraA_1);
// ...
// ...
addChild(LetraB_1);
ABClist.push(LetraB_1);
// ...
// Finally.
function eliminarhijos(e:MouseEvent):void
{
// Iterate over the listed letters.
for each (var aLetter:DisplayObject in ABClist)
{
// Removes each of the listed letters, one by one.
removeChild(aLetter);
}
// Clear the list.
ABClist.length = 0;
}
方式№3。标记。如果出于任何不可能的原因,上述任何一项都不适合您,您可以以特定方式命名这些字母,以便将它们从其他对象中过滤掉。
// Inside the letter creation methods.
// ...
addChild(LetraA_1);
LetraA_1.name = "LETTER";
// ...
// ...
addChild(LetraB_1);
LetraB_1.name = "LETTER";
// ...
// Finally.
function eliminarhijos(e:MouseEvent):void
{
// Iterate over all children. Backward loop, because if you
// remove something, the whole body of children shifts down.
for (var i:int = numChildren - 1; i >= 0; i--)
{
var aChild:DisplayObject = getChildAt(i);
// So you have a criteria to figure out if it is a letter or not.
if (aChild.name == "LETTER")
{
removeChild(aChild);
}
}
}
方式№4。进入诡异。如果以上都不适合你,还有一种方法可以将字母与那里的其他对象分开。
// You need to list all the classes of your letters here.
var LetterBox:Array = [letraA, letraB];
// The clean up.
function eliminarhijos(e:MouseEvent):void
{
// Iterate over all children. Backward loop, because if you
// remove something, the whole body of children shifts down.
for (var i:int = numChildren - 1; i >= 0; i--)
{
var aChild:DisplayObject = getChildAt(i);
// Now iterate over all possible letter classes
// to figure out if the given child belongs to any of them.
for each (var aClass:Class in LetterBox)
{
// Match criteria.
if (aChild is aClass)
{
removeChild(aChild);
break;
}
}
}
}