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查找 XML 条目号
原标题:Find XML entry number

我需要获得一个条目的节点编号,但我只有LOG_ID。怎么找到这个号码?

<LOG>
<ENTRY LOG_ID="01042012"/>
<ENTRY LOG_ID="03052012"/>
<ENTRY LOG_ID="09052012"/>
</LOG>

谢谢,乌利

最佳回答

使用描述为 < a href="的 E4X 处理 http://help.adobe.com/en_US/ActionScript/3.0_ProgramingAS3/WS5b3cc516d4ff351e63e3d118a9b90204-7e72.html" rel="no follow" >这里 和 < a href="http://wso2.org/project/mashup/0.2/docs/e4xquickstart.html" rel="nofollow" 开始的文档:

var myXML:XML = 
 <LOG>
  <ENTRY LOG_ID="01042012"/>
  <ENTRY LOG_ID="03052012"/>
  <ENTRY LOG_ID="09052012"/>
 </LOG>

trace( myXML.ENTRY.(@LOG_ID==09052012).childIndex() ); /* retrieve entire node */

您也可以在 XML 对象中保存此节点的引用 :

 var index:int = myXML.ENTRY.(@LOG_ID==09052012).childIndex();

注意 : childind 函数( 和其他一些函数) 工作在单个节点上。 但是, 如果您输入示例有多个节点, 且与您在检索时使用的属性值相同, 您将得到一个节点列表( 即 < code> XMList ), 而不是一个节点 。 现在, 为了查找这些孩子的索引, 您需要做以下操作 :

for each ( var selectedNode in myXML.ENTRY.(@LOG_ID==09052012) )
    trace( selectedNode.childIndex() );

您总是可以通过下列方式检查您的 E4X 查询是否返回了列表 :

var candidates:XMLList = myXML.ENTRY.(@LOG_ID==09052012) as XMLList;
if (candidates != null) { // a list 
        // do something ...
}
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