Question

如何最有效地通过 Java的所有DOM分子加以利用?

与此类似,但对于目前<代码>org.w3c.dom.Document上的每一个单项OM要素而言?

for(Node childNode = node.getFirstChild(); childNode!=null;){
    Node nextChild = childNode.getNextSibling();
    // Do something with childNode, including move or delete...
    childNode = nextChild;
}

Answer 1

基本来说,你有两种方式来振兴所有要素:

1. Using recursion (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. 避免再侵入<>/strong> ,使用<条码>的植被要素ByTagName( 和*作为参数:

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,这些途径是有效的。

Answer 2

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

改变

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

提高效率。

second万纳答案的第二种方式可能是最好的,因为它倾向于使用一种更固定、可预测的记忆模式。

Answer 3

I also stumbled over this problem recently. Here is my solution. I wanted to avoid recursion, so I used a while loop.

Because of the adds and removes in arbitrary places on the list, I went with the LinkedList implementation.

/* traverses tree starting with given node */
  private static List<Node> traverse(Node n)
  {
    return traverse(Arrays.asList(n));
  }

  /* traverses tree starting with given nodes */
  private static List<Node> traverse(List<Node> nodes)
  {
    List<Node> open = new LinkedList<Node>(nodes);
    List<Node> visited = new LinkedList<Node>();

    ListIterator<Node> it = open.listIterator();
    while (it.hasNext() || it.hasPrevious())
    {
      Node unvisited;
      if (it.hasNext())
        unvisited = it.next();
      else
        unvisited = it.previous();

      it.remove();

      List<Node> children = getChildren(unvisited);
      for (Node child : children)
        it.add(child);

      visited.add(unvisited);
    }

    return visited;
  }

  private static List<Node> getChildren(Node n)
  {
    List<Node> children = asList(n.getChildNodes());
    Iterator<Node> it = children.iterator();
    while (it.hasNext())
      if (it.next().getNodeType() != Node.ELEMENT_NODE)
        it.remove();
    return children;
  }

  private static List<Node> asList(NodeList nodes)
  {
    List<Node> list = new ArrayList<Node>(nodes.getLength());
    for (int i = 0, l = nodes.getLength(); i < l; i++)
      list.add(nodes.item(i));
    return list;
  }

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