Java：最有效的方法来遍历org.w3c.dom.Document中的所有元素？ [英] Java: Most efficient method to iterate over all elements in a org.w3c.dom.Document?

问题描述

这样的东西，但是对于当前上的每一个DOM元素，最有效的方式是遍历Java中的所有DOM元素（Node childNode = node.getFirstChild（）;）$（$）


$ b $ childNode！= null;）{
Node nextChild = childNode.getNextSibling（）;
//使用childNode执行某些操作，包括移动或删除...
childNode = nextChild;
}


解决方案

基本上你有两种方法迭代所有元素：

1。使用递归（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）{
 //使用当前节点而不是System.out执行某些操作
 System.out.println（node .getNodeName（））; 
 
 NodeList nodeList = node.getChildNodes（）; （int i = 0; i< nodeList.getLength（）; i ++）{
 Node currentNode = nodeList.item（i）; 
 
 if（currentNode.getNodeType（）== Node.ELEMENT_NODE）{
 //为所有子元素调用此方法，即Element 
 doSomething（currentNode）; 
} 
} 
} 
  

2。使用 getElementsByTagName（）方法使用 * 作为参数避免递归：

  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（*）; （int i = 0; i< nodeList.getLength（）; i ++）{
 Node node = nodeList.item（i）; 
 
 if（node.getNodeType（）== Node.ELEMENT_NODE）{
 //使用当前元素执行某些操作
 System.out.println（node.getNodeName（））; 
} 
} 
} 
  

我认为这两种方式都是有效。

希望这有帮助。

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

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;
}

解决方案

Basically you have two ways to iterate over all elements:

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. 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());
        }
    }
}

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

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