Java:迭代 org.w3c.dom.Document 中所有元素的最有效方法? [英] Java: Most efficient method to iterate over all elements in a org.w3c.dom.Document?

问题描述

在 Java 中迭代所有 DOM 元素的最有效方法是什么?

What is the most efficient way to iterate through all DOM elements in Java?

类似的东西但是对于当前 org.w3c.dom.Document 上的每个 DOM 元素?

Something like this but for every single DOM elements on current 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;
}

推荐答案

基本上你有两种方法来迭代所有元素:

Basically you have two ways to iterate over all elements:

1.使用递归(我认为最常用的方式):

1. Using recursion (the most common way I think):

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() 方法和 * 作为参数避免递归:

2. Avoiding recursion using getElementsByTagName() method with * as parameter:

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());
        }
    }
}

我认为这两种方式都很有效.
希望这会有所帮助.

I think these ways are both efficient.
Hope this helps.

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