请问：dom4j怎么遍历某一个节点（节点是通过属性名指定）下的子节点

Question

<?xml version='1.0' encoding='utf-8'?>
<tableList>
 <table name="dd">
  <column key="d" value="avc">bac</column>
 </table>
<table name="AA">
  <column key="d" value="avc">bac</column>
</table>

<table name="cc">
  <column key="d" value="avc">bac</column>
</table>

</tableList>

通过table 的name属性遍历column子节点，先谢谢大家了，

Answer 1

亲，这样试试~

String content = "xml文件字符串";
Document doc = DocumentHelper.parseText(content);
List tableList = doc.selectNodes("//tableList/table");
for (Object tableObj : tableList ) {
   Element table = (Element) tableObj;
   if(null != table){
      String tableName = table.attributeValue("name");
      List columnList = table.getDocument().selectNodes("//table/column");
      for(Object columnObj : columnList){
         Element column= (Element) tableObj;
         String key = column.attributeValue("key");
         String value = column.attributeValue("value");
         String text = column.getTextTrim();
      )
   }
}

Answer 2

最简单的办法 

SAXReader reader = new SAXReader();
		Element root = reader.read("fileName.xml").getDocument()
				.getRootElement();
		List<Element> nodes = root.elements("table");
		for (Element node : nodes) {
			if ("xxxx".equals(node.attributeValue("name"))) {
				Element column = node.element("column");
				//some code
			}
		}

另外 你还可以用xpath的方式来解决

Answer 3

亲，这样试试~

String content = "xml文件字符串";
Document doc = DocumentHelper.parseText(content);
List tableList = doc.selectNodes("//tableList/table");
for (Object tableObj : tableList ) {
   Element table = (Element) tableObj;
   if(null != table){
      String tableName = table.attributeValue("name");
      List columnList = table.getDocument().selectNodes("//table/column");
      for(Object columnObj : columnList){
         Element column= (Element) tableObj;
         String key = column.attributeValue("key");
         String value = column.attributeValue("value");
         String text = column.getTextTrim();
      )
   }
}

Answer 4

最简单的办法 

SAXReader reader = new SAXReader();
		Element root = reader.read("fileName.xml").getDocument()
				.getRootElement();
		List<Element> nodes = root.elements("table");
		for (Element node : nodes) {
			if ("xxxx".equals(node.attributeValue("name"))) {
				Element column = node.element("column");
				//some code
			}
		}

另外 你还可以用xpath的方式来解决