问题
Is it possible to sort an XMLList? All the examples I can find on it create a new XMLListCollection like this:
MyXMLListCol = new XMLListCollection(MyXMLList);
I don't think the XMLListCollection in this case has any reference to the XMLList so sorting it would leave my XMLList unsorted, is this correct?
How can I sort the XMLList directly?
Thanks ~Mike
回答1:
So I finally got my search terms altered enough I actually churned up an answer to this. Using the technique I got from here: http://freerpad.blogspot.com/2007/07/more-hierarchical-sorting-e4x-xml-for.html
I was able to come up with this:
public function sortXMLListByAttribute(parentNode:XML,xList:XMLList,attr:String):void{
//attr values must be ints
var xListItems:int = xList.length();
if(xListItems !=0){
var sortingArray:Array = new Array();
var sortAttr:Number = new Number();
for each (var item:XML in xList){
sortAttr = Number(item.attribute(attr));
if(sortingArray.indexOf(sortAttr)==-1){
sortingArray.push(sortAttr);
}
//piggy back the removal, just have to remove all of one localName without touching items of other localNames
delete parentNode.child(item.localName())[0];
}
if( sortingArray.length > 1 ) {
sortingArray.sort(Array.NUMERIC);
}
var sortedList:XMLList = new XMLList();
for each(var sortedAttr:Number in sortingArray){
for each (var item2:XML in xList){
var tempVar:Number = Number(item2.attribute(attr));
if(tempVar == sortedAttr){
sortedList += item2
}
}
}
for each(var item3:XML in sortedList){
parentNode.appendChild(item3);
}
}
}
Works pretty fast and keeps my original XML variable updated. I know I may be reinventing the wheel just to not use an XMLListCollection, but I think the ability to sort XML and XMLLists can be pretty important
回答2:
While there is no native equivalent to the Array.sortOn function, it is trivial enough to implement your own sorting algorithm:
// Bubble sort.
// always initialize variables -- it save memory.
var ordered:Boolean = false;
var l:int = xmlList.length();
var i:int = 0;
var curr:XML = null;
var plus:XML = null;
while( !ordered )
{
// Assume that the order is correct
ordered = true;
for( i = 0; i < l; i++ )
{
curr = xmlList[ i ];
plus = xmlList[ i + 1 ];
// If the order is incorrect, swap and set ordered to false.
if( Number( curr.@order ) < Number( plus.@order ) )
{
xmlList[ i ] = plus;
xmlList[ i + 1 ] = curr;
ordered = false;
}
}
}
but, realistically, it is far easier and less buggy to use XMLListCollection. Further, if someone else is reading your code, they will find it easier to understand. Please do yourself a favor and avoid re-inventing the wheel on this.
来源:https://stackoverflow.com/questions/1869964/flex-newbie-xmllist-question-sorting-xml-and-xmllist