Question

我的目标是将每个dom元素（Node.ELEMENT_NODE）包装在标有org.w3c.dom.Document的当前<something style="background-color:red"></something>上。

public static void main(String[] args){
    org.w3c.dom.DOMDocument doc;
    paintAllNodes(doc, 0);
}

public static void paintAllNodes(Node node, int level) {
    // Process node

    // If there are any children, visit each one
    NodeList list = node.getChildNodes();
    for (int i=0; i<list.getLength(); i++) {
        // Get child node
        Node childNode = list.item(i);        

        // Visit child node
        paintAllNodes(childNode, level+1);
    }
}

Answer 1

解决此类问题的最简单方法之一（就像任何XML转换一样）是使用XSLT 。

此XSLT转换：

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

 <xsl:template match="*">
 <something style="background-color:red">
   <xsl:copy>
    <xsl:copy-of select="@*"/>
    <xsl:apply-templates select="node()"/>
   </xsl:copy>
  </something>
 </xsl:template>
</xsl:stylesheet>

应用于任何XML文档时，例如：

<nums>
  <num>01</num>
  <num>02</num>
  <num>03</num>
  <num>04</num>
  <num>05</num>
  <num>06</num>
  <num>07</num>
  <num>08</num>
  <num>09</num>
  <num>10</num>
</nums>

生成想要的正确结果：

<something style="background-color:red">
   <nums>
      <something style="background-color:red">
         <num>01</num>
      </something>
      <something style="background-color:red">
         <num>02</num>
      </something>
      <something style="background-color:red">
         <num>03</num>
      </something>
      <something style="background-color:red">
         <num>04</num>
      </something>
      <something style="background-color:red">
         <num>05</num>
      </something>
      <something style="background-color:red">
         <num>06</num>
      </something>
      <something style="background-color:red">
         <num>07</num>
      </something>
      <something style="background-color:red">
         <num>08</num>
      </something>
      <something style="background-color:red">
         <num>09</num>
      </something>
      <something style="background-color:red">
         <num>10</num>
      </something>
   </nums>
</something>

注意：作为练习，如何在Java代码中启动XSLT转换：）

Answer 2

这会将Node.ELEMENT_NODE中的每个org.w3c.dom.Document换成<something>个标记：

public static void paintAllNodes(Document doc) {
    // build list of Node.ELEMENT_NODE to process
    List<Node> nodes = new ArrayList<Node>();
    NodeList list = doc.getElementsByTagName("*");
    for (int i = 0; i < list.getLength(); i++) {
        Node node = list.item(i);
        if (node.getNodeType() == Node.ELEMENT_NODE) {
            nodes.add(node);
        }
    }

    // iterate through each node and wrap with <something> tag
    for (Node node : nodes) {
        // remember the next sibling for inserting at end
        Node nextSibling = node.getNextSibling();

        // remember the parent and remove this node from it
        Node parent = node.getParentNode();
        parent.removeChild(node);

        // create <something> element and attach node
        Element element = doc.createElement("something");
        NamedNodeMap attributes = element.getAttributes();
        Attr attr = doc.createAttribute("style");
        attr.setNodeValue("background-color:red");
        attributes.setNamedItem(attr);
        element.appendChild(node);

        // insert new element where the node was 
        parent.insertBefore(element, nextSibling);
    }
}

如果要排除任何节点，只需在第一个for循环中过滤掉它们。

Answer 3

我想建议一个递归解决方案，它使用Node#replaceChild方法用新标签替换节点：

public static void paintAllNodes(Node node) {
    if (node.getNodeType() == Node.ELEMENT_NODE) {
        Element somethingElement = node.getOwnerDocument().createElement("something");
        somethingElement.setAttribute("style", "background-color:red");
        node.getParentNode().replaceChild(somethingElement, node);
        somethingElement.appendChild(node);
        NodeList nodeList = node.getChildNodes();
        for (int i = 0; i < nodeList.getLength(); i++) {
            paintAllNodes(nodeList.item(i));
        }
    }
}

这是我的主要内容：

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"));
    paintAllNodes(document.getDocumentElement());

    TransformerFactory transformerFactory = TransformerFactory.newInstance();
    Transformer transformer = transformerFactory.newTransformer();
    DOMSource source = new DOMSource(document);
    StreamResult result = new StreamResult(System.out);
    transformer.transform(source, result);
}

我用这个xml测试了它：

<html>
    <head>
        <title>title</title>
    </head>
    <body>
    <h1>title</h1>
    <div>test</div>
    </body>
</html>

我的主要打印出这个新的xml，这似乎是你想要的：

<?xml version="1.0" encoding="UTF-8"?><something style="background-color:red"><html>
    <something style="background-color:red"><head>
        <something style="background-color:red"><title>title</title></something>
    </head></something>
    <something style="background-color:red"><body>
    <something style="background-color:red"><h1>title</h1></something>
    <something style="background-color:red"><div>test</div></something>
    </body></something>
</html></something>

希望这有帮助。

Answer 4

作为非答案;-)我建议使用CSS ... something {background:red} 显然，这只有在你使用CSS时才有意义。

Answer 5

对于任何给定节点“节点”，文档“文档”来自此类应该有效。

Node parent = node.getParent();
Node nextSibling = node.getNextSibling();

parent.removeChild(node);

Element something = document.createElement("something");
NamedNodeMap atts = something.getAttributes();
Attr att = document.createAttribute("style");
att.setNodeValue("background-color:red;");
atts.setNamedItem(att);
something.addChild(node);
parent.insertBefore(something, nextSibling);

Java：如何用org.w3c.dom中的<sometag>包装所有元素？</sometag>

5 个答案: