使用nodeList创建XML文档

Question

我需要使用NodeList创建XML文档对象。有人能帮我做到这一点。我已经向您展示的代码及以下

import 
javax.xml.parsers.DocumentBuilderFactory; 
import javax.xml.xpath.*; import 
org.w3c.dom.*; 

public class ReadFile { 

    public static void main(String[] args) { 
     String exp = "/configs/markets"; 
     String path = "testConfig.xml"; 
     try { 
      Document xmlDocument = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(path); 
      XPath xPath = XPathFactory.newInstance().newXPath(); 
      XPathExpression xPathExpression = xPath.compile(exp); 
      NodeList nodes = (NodeList) 
       xPathExpression.evaluate(xmlDocument, 
             XPathConstants.NODESET); 

     } catch (Exception ex) { 
      ex.printStackTrace(); 
     } 
    } 
} 

XML文件的XML如下所示

<configs> 
    <markets> 
     <market> 
      <name>Real</name> 
     </market> 
     <market> 
      <name>play</name> 
     </market> 
    </markets> 
</configs> 

在此先感谢..

Answer 1

你应该做的是这样的：

• 创建新org.w3c.dom.Document newXmlDoc您存储节点在NodeList，
• 您创建一个新的根元素，将其追加到newXmlDoc
• 然后，对每个节点n在NodeList，你导入newXmlDocn，然后追加n为root

这孩子是代码：

public static void main(String[] args) { 
    String exp = "/configs/markets/market"; 
    String path = "src/a/testConfig.xml"; 
    try { 
     Document xmlDocument = DocumentBuilderFactory.newInstance() 
       .newDocumentBuilder().parse(path); 

     XPath xPath = XPathFactory.newInstance().newXPath(); 
     XPathExpression xPathExpression = xPath.compile(exp); 
     NodeList nodes = (NodeList) xPathExpression. 
       evaluate(xmlDocument, XPathConstants.NODESET); 

     Document newXmlDocument = DocumentBuilderFactory.newInstance() 
       .newDocumentBuilder().newDocument(); 
     Element root = newXmlDocument.createElement("root"); 
     newXmlDocument.appendChild(root); 
     for (int i = 0; i < nodes.getLength(); i++) { 
      Node node = nodes.item(i); 
      Node copyNode = newXmlDocument.importNode(node, true); 
      root.appendChild(copyNode); 
     } 

     printTree(newXmlDocument); 
    } catch (Exception ex) { 
     ex.printStackTrace(); 
    } 
} 

public static void printXmlDocument(Document document) { 
    DOMImplementationLS domImplementationLS = 
     (DOMImplementationLS) document.getImplementation(); 
    LSSerializer lsSerializer = 
     domImplementationLS.createLSSerializer(); 
    String string = lsSerializer.writeToString(document); 
    System.out.println(string); 
} 

输出是：

<?xml version="1.0" encoding="UTF-16"?> 
<root><market> 
      <name>Real</name> 
     </market><market> 
      <name>play</name> 
     </market></root> 

一些注意事项：

• 我已经改变了exp到/configs/markets/market，因为我怀疑你要复制的market元素，而不是单一的markets元素
• 为printXmlDocument，我使用了有趣的代码在此answer

我希望这有帮助。

---

如果你不希望创建一个新的根元素，那么你可以使用你原来的XPath表达式，它返回一个NodeList由单个节点（请记住，你的XML必须有一个根元素），您可以直接将其添加到新的XML文档中。

请参见下面的代码，在那里我评论从上面的代码行：

public static void main(String[] args) { 
    //String exp = "/configs/markets/market/"; 
    String exp = "/configs/markets"; 
    String path = "src/a/testConfig.xml"; 
    try { 
     Document xmlDocument = DocumentBuilderFactory.newInstance() 
       .newDocumentBuilder().parse(path); 

     XPath xPath = XPathFactory.newInstance().newXPath(); 
     XPathExpression xPathExpression = xPath.compile(exp); 
     NodeList nodes = (NodeList) xPathExpression. 
     evaluate(xmlDocument,XPathConstants.NODESET); 

     Document newXmlDocument = DocumentBuilderFactory.newInstance() 
       .newDocumentBuilder().newDocument(); 
     //Element root = newXmlDocument.createElement("root"); 
     //newXmlDocument.appendChild(root); 
     for (int i = 0; i < nodes.getLength(); i++) { 
      Node node = nodes.item(i); 
      Node copyNode = newXmlDocument.importNode(node, true); 
      newXmlDocument.appendChild(copyNode); 
      //root.appendChild(copyNode); 
     } 

     printXmlDocument(newXmlDocument); 
    } catch (Exception ex) { 
     ex.printStackTrace(); 
    } 
} 

这会给你以下的输出：

<?xml version="1.0" encoding="UTF-16"?> 
<markets> 
     <market> 
      <name>Real</name> 
     </market> 
     <market> 
      <name>play</name> 
     </market> 
    </markets> 

Answer 2

您可以尝试Document的adoptNode()方法。

也许你需要遍历你的NodeList。您可以通过nodeList.item(i)访问个人Nodes。

如果你想包装在Element搜索结果，你可以使用的createElement（）的新创建的Document和appendChild()Element