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

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在Java中遍历所有DOM元素的最有效方法是什么?

像这样的东西,但对目前的每一个DOM元素org.w3c.dom.Document

for(Node childNode = node.getFirstChild(); childNode!=null;){
    Node nextChild = childNode.getNextSibling();
    // Do something with childNode, including move or delete...
    childNode = nextChild;
}
提问于
用户回答回答于

基本上你有两种方法遍历所有元素:

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.避免使用getElementsByTagName()带有*as参数的方法递归

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());
        }
    }
}
用户回答回答于

for (int i = 0; i < nodeList.getLength(); i++)

改成

for (int i = 0, len = nodeList.getLength(); i < len; i++)

更高效。第二种方法可能是最好的,因为它倾向于使用更扁平,可预测的内存模型。

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