Java：最有效的方法来遍历org.w3c.dom.Document中的所有元素？

基本上，您可以通过两种方式遍历所有元素：

1.使用递归 （我认为是最常见的方式）：

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

getElementsByTagName()

*

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());        }    }}

我认为这些方式都很有效。
希望这可以帮助。