基本上你有两种方法来遍历所有元素：
1.使用递归(我认为最常见的方式)：
public static void main(String[] args) throws SAXException, IOException,
ParserConfigurationException, TransformerException {
DocumentBuilderFactory docBuilderFactory = DocumentBuilderFactory
.newInstance();
DocumentBuilder docBuilder = docBuilderFactory.newDocumentBuilder();
Document document = docBuilder.parse(new File("document.xml"));
doSomething(document.getDocumentElement());
}
public static void doSomething(Node node) {
// do something with the current node instead of System.out
System.out.println(node.getNodeName());
NodeList nodeList = node.getChildNodes();
for (int i = 0; i < nodeList.getLength(); i++) {
Node currentNode = nodeList.item(i);
if (currentNode.getNodeType() == Node.ELEMENT_NODE) {
//calls this method for all the children which is Element
doSomething(currentNode);
}
}
}
2.使用带*作为参数的getElementsByTagName()方法避免递归：
public static void main(String[] args) throws SAXException, IOException,
ParserConfigurationException, TransformerException {
DocumentBuilderFactory docBuilderFactory = DocumentBuilderFactory
.newInstance();
DocumentBuilder docBuilder = docBuilderFactory.newDocumentBuilder();
Document document = docBuilder.parse(new File("document.xml"));
NodeList nodeList = document.getElementsByTagName("*");
for (int i = 0; i < nodeList.getLength(); i++) {
Node node = nodeList.item(i);
if (node.getNodeType() == Node.ELEMENT_NODE) {
// do something with the current element
System.out.println(node.getNodeName());
}
}
}
我认为这些方法都是高效的。希望这可以帮助。