Flex tree expand slowly/lazy 超过30000个节点时加载
Flex tree expand slowly/lazy loading when there are more than 30000 nodes
我想以 UI 始终响应的方式打开 Flex mx Tree 的所有节点。通常当节点数量有限时,这段代码就可以正常工作。
public function expandAllNodes():void {
for (var i:int = 0; i < this.dataProvider.length; i++) {
expandChildrenOf(this.dataProvider.getItemAt(i), true);
}
}
也许我需要callLater而不是直接调用函数
public function expandAllNodes():void {
for (var i:int = 0; i < this.dataProvider.length; i++) {
calllater(expandChildrenOf,[this.dataProvider.getItemAt(i), true]);
}
}
但这也不起作用。
在这种情况下使用 callLater 无济于事。来自documentation,
The callLater method queues an operation to be performed for the next screen refresh, rather than in the current update.
在您的循环中使用 callLater 只会将 all expandChildrenOf() 调用推送到下一个屏幕刷新 - 这仍然会在单个更新周期中投入太多工作。
相反,您可以将循环分布在多个帧上并限制每帧完成的工作:
private var processingLimit:int = 10; //experiment to get a reasonable value for this
private var totalNodes:int = 0;
private var nodesProcessed:int = 0;
public function expandAllNodes():void {
this.totalNodes = this.dataProvider.length;
this.addEventListener(Event.ENTER_FRAME, expandNextNode);
}
private function expandNextNode(e:Event):void {
var numToProcess = Math.min(processingLimit + nodesProcessed, totalNodes);
for (var i:int = nodesProcessed; i < numToProcess; i++) {
nodesProcessed++;
expandChildrenOf(this.dataProvider.getItemAt(i), true);
}
if (numToProcess == totalNodes) {
this.removeEventListener(Event.ENTER_FRAME, expandNextNode);
}
}
n.b。确保 expandChildrenOf
不会递归地打开子项的子项——如果是这样,那么您仍然可以根据对 expandChildrenOf(1);
!
的单次调用来展开整个树。
我就是这样解决问题的
private const PROCESSING_LIMIT:int = 25;
public var _expandNodesStack:Vector.<Object>;
private function expandNextNode(e:Event):void {
for (var i:int = 0; i < PROCESSING_LIMIT; i++) {
var item:* = _expandNodesStack.pop();
if (item)
expandItem(item, true);
else
break;
}
if (_expandNodesStack.length == 0) {
this.removeEventListener(Event.ENTER_FRAME, expandNextNode);
}
}
public function expandAllNodes():void {
if(this.dataProvider == null || this.dataProvider.length <= 0) return;
if (_expandNodesStack != null && _expandNodesStack.length > 0) {
//already expanding so abort this operation
return;
}
if (_expandNodesStack == null) _expandNodesStack = new <Object>[];
iterateAndPushToStack(this.dataProvider.getItemAt(0));
_expandNodesStack.reverse();
this.addEventListener(Event.ENTER_FRAME, expandNextNode);
}
private function iterateAndPushToStack(item:Object):void {
if (iterator == null)
return;
// if it is not a branch item there's nothing to do
if (item != null && item.children is ArrayCollection && item.children.length > 0) {
_expandNodesStack.push(item);
var childItems:ArrayCollection;
childItems = item.children;
if (childItems) {
for each (var object:Object in childItems) {
iterateAndPushToStack(object);
}
}
}
}
我想以 UI 始终响应的方式打开 Flex mx Tree 的所有节点。通常当节点数量有限时,这段代码就可以正常工作。
public function expandAllNodes():void {
for (var i:int = 0; i < this.dataProvider.length; i++) {
expandChildrenOf(this.dataProvider.getItemAt(i), true);
}
}
也许我需要callLater而不是直接调用函数
public function expandAllNodes():void {
for (var i:int = 0; i < this.dataProvider.length; i++) {
calllater(expandChildrenOf,[this.dataProvider.getItemAt(i), true]);
}
}
但这也不起作用。
在这种情况下使用 callLater 无济于事。来自documentation,
The callLater method queues an operation to be performed for the next screen refresh, rather than in the current update.
在您的循环中使用 callLater 只会将 all expandChildrenOf() 调用推送到下一个屏幕刷新 - 这仍然会在单个更新周期中投入太多工作。
相反,您可以将循环分布在多个帧上并限制每帧完成的工作:
private var processingLimit:int = 10; //experiment to get a reasonable value for this
private var totalNodes:int = 0;
private var nodesProcessed:int = 0;
public function expandAllNodes():void {
this.totalNodes = this.dataProvider.length;
this.addEventListener(Event.ENTER_FRAME, expandNextNode);
}
private function expandNextNode(e:Event):void {
var numToProcess = Math.min(processingLimit + nodesProcessed, totalNodes);
for (var i:int = nodesProcessed; i < numToProcess; i++) {
nodesProcessed++;
expandChildrenOf(this.dataProvider.getItemAt(i), true);
}
if (numToProcess == totalNodes) {
this.removeEventListener(Event.ENTER_FRAME, expandNextNode);
}
}
n.b。确保 expandChildrenOf
不会递归地打开子项的子项——如果是这样,那么您仍然可以根据对 expandChildrenOf(1);
!
我就是这样解决问题的
private const PROCESSING_LIMIT:int = 25;
public var _expandNodesStack:Vector.<Object>;
private function expandNextNode(e:Event):void {
for (var i:int = 0; i < PROCESSING_LIMIT; i++) {
var item:* = _expandNodesStack.pop();
if (item)
expandItem(item, true);
else
break;
}
if (_expandNodesStack.length == 0) {
this.removeEventListener(Event.ENTER_FRAME, expandNextNode);
}
}
public function expandAllNodes():void {
if(this.dataProvider == null || this.dataProvider.length <= 0) return;
if (_expandNodesStack != null && _expandNodesStack.length > 0) {
//already expanding so abort this operation
return;
}
if (_expandNodesStack == null) _expandNodesStack = new <Object>[];
iterateAndPushToStack(this.dataProvider.getItemAt(0));
_expandNodesStack.reverse();
this.addEventListener(Event.ENTER_FRAME, expandNextNode);
}
private function iterateAndPushToStack(item:Object):void {
if (iterator == null)
return;
// if it is not a branch item there's nothing to do
if (item != null && item.children is ArrayCollection && item.children.length > 0) {
_expandNodesStack.push(item);
var childItems:ArrayCollection;
childItems = item.children;
if (childItems) {
for each (var object:Object in childItems) {
iterateAndPushToStack(object);
}
}
}
}