我如何从每个父节点分别获取子节点? [英] how I get child node from each parent node separately?
问题描述
我有一些Data Xml ..
< main>
< TabNavigator x =27y =11width =455height =376id =ghbackgroundColor =#A4B6E9>
< NavigatorContent width =100%height =100%label =Clientid =clientTab>< / NavigatorContent>
< TitleWindow x =521y =84width =377height =234>
< DataGrid x =0y =0width =375height =163borderVisible =trueid =details>
<栏>
< ArrayList>
< GridColumn dataField =Nameid =arrayName/>< GridColumn dataField =AddressheaderText =Address/>
< GridColumn dataField =Phone_NumberheaderText =Phone_Number/>
< / ArrayList>
< / columns>
< / DataGrid>
< / TitleWindow>
< / main>
我使用以下代码来检索给定XML的子名称。 b
private function urlLdr_complete(event:Event):void {
var xmlData:XML = new XML(URLLoader(event.currentTarget).data);
(var t:xmlData.children()中的XML)
{
Alert.show(t.Name);
}
但是我只得到2个孩子(TabNavigator和TitleWindow)其他孩子在每个父节点?我想每个家长分开的孩子。我怎样才能得到它?
您需要使用递归函数来遍历树。使用trace()而不是alert():
private function urlLdr_complete(event:Event):void
{
var xmlData:XML = new XML(URLLoader(event.currentTarget).data);
showNodeName(xmlData);
$ b $ private函数showNodeName($ node:XML):void
{
//跟踪当前节点
trace($ node.name ));
if($ node.hasChildNodes)
{
for each(var child:XML in $ node.children())
{
//递归调用此函数每个孩子
showNodeName(child);
$ p $或者使用E4X后代()函数:
pre $ code私有函数urlLdr_complete(event:Event):void
{
var xmlData: XML = new XML(URLLoader(event.currentTarget).data);
//跟踪根节点:$ b $ b trace(xmlData.name());
//并跟踪其所有后代:
(var child:xmlData.descendants()中的XML)
{
trace(child.name());
$ p $都应该产生相同的结果:$ / b
$ TabNavigator
NavigatorContent
NavigatorContent
TitleWindow
DataGrid
列
ArrayList
GridColumn $ b $ GridColumn
GridColumn
Button
$ / code>
我没有测试,但我期望内置的后代()函数更有效率。
I have some Data Xml..
<main>
<TabNavigator x="27" y="11" width="455" height="376" id="gh" backgroundColor="#A4B6E9">
<NavigatorContent width="100%" height="100%" label="Client" id="clientTab"></NavigatorContent>
<NavigatorContent width="100%" height="100%" label="Admin" id="adminTab"></NavigatorContent></TabNavigator>
<TitleWindow x="521" y="84" width="377" height="234">
<DataGrid x="0" y="0" width="375" height="163" borderVisible="true" id="details">
<columns>
<ArrayList>
<GridColumn dataField="Name" id="arrayName"/><GridColumn dataField="Address" headerText="Address"/>
<GridColumn dataField="Phone_Number" headerText="Phone_Number"/>
</ArrayList>
</columns>
</DataGrid>
<Button x="139" y="167" height="28" label="Export"/>
</TitleWindow>
</main>
I use following code for retrieving the child names of given XML..
private function urlLdr_complete(event:Event):void{
var xmlData:XML=new XML(URLLoader(event.currentTarget).data);
for each (var t:XML in xmlData.children())
{
Alert.show(t.Name);
}
But I only get 2 children(TabNavigator and TitleWindow).How do I get the other children in each parent node? I want separate children for each parent. How can I get it?
You need to use a recursive function to walk down the tree. Using trace() instead of alert():
private function urlLdr_complete(event:Event):void
{
var xmlData:XML=new XML(URLLoader(event.currentTarget).data);
showNodeName(xmlData);
}
private function showNodeName($node:XML):void
{
// Trace the current node
trace($node.name());
if($node.hasChildNodes)
{
for each (var child:XML in $node.children())
{
// Recursively call this function on each child
showNodeName(child);
}
}
}
Or, use the E4X descendants() function:
private function urlLdr_complete(event:Event):void
{
var xmlData:XML=new XML(URLLoader(event.currentTarget).data);
// Trace the root node:
trace(xmlData.name());
// And trace all its descendants:
for each(var child:XML in xmlData.descendants())
{
trace(child.name());
}
}
Both should produce identical outcomes:
main
TabNavigator
NavigatorContent
NavigatorContent
TitleWindow
DataGrid
columns
ArrayList
GridColumn
GridColumn
GridColumn
Button
I haven't tested but I would expect the built-in descendants() function to be more efficient.
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