使用nodeList创建XML文档 [英] Create XML document using nodeList

问题描述

我需要使用NodeList创建一个XML文档对象。有些人可以帮助我做到这一点。我已经向您显示了代码，并在下面的

  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文件显示如下

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

提前感谢

你应该这么做：

• 你创建一个新的 org。 w3c.dom.Document newXmlDoc 将节点存储在 NodeList 中，


• ，您将创建一个新的根元素，并将其附加到 newXmlDoc


• 然后，对于每个节点 n NodeList 中，在 newXmlDoc 中导入 n 然后你将 n 作为 root

$ b的孩子
$ b

这是代码：

  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）; 
 
文档newXmlDocument = DocumentBuilderFactory.newInstance（）
 .newDocumentBuilder（）。newDocument（）; 
元素root = newXmlDocument.createElement（root）; 
 newXmlDocument.appendChild（root）; （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 =1.0encoding =UTF-16？> 
< root>< market> 
< name> Real< / name> 
< / market>< market> 
< name> play< / name> 
< / market>< / root> 
  

一些注释：

• 我已将 exp 更改为 / configs / markets / market ，因为我怀疑你要复制市场元素，而不是单个市场元素


• code> printXmlDocument ，我使用了这个 answer

I希望这有帮助。

---

如果不想创建新的根元素，那么可以使用原来的XPath表达式返回一个由单个节点组成的 NodeList （请记住，您的XML必须有一个根元素），您可以直接添加到新的XML文档中。 / p>

请参阅以下代码，我在上面的代码中注释了行：

  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）; 
 
文档newXmlDocument = DocumentBuilderFactory.newInstance（）
 .newDocumentBuilder（）。newDocument（）; 
 //元素root = newXmlDocument.createElement（root）; 
 //newXmlDocument.appendChild(root）; （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（）; 
} 
} 
  

这将给你以下输出：

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

I need to create a XML Document object using the NodeList. Can some one pls help me to do this. I have shown you the code and the xml below

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 file is shown below

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

Thanks in advance..

解决方案

You should do it like this:

• you create a new org.w3c.dom.Document newXmlDoc where you store the nodes in your NodeList,
• you create a new root element, and append it to newXmlDoc
• then, for each node n in your NodeList, you import n in newXmlDoc, and then you append n as a child of root

Here is the code:

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);
}

The output is:

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

Some notes:

• I've changed exp to /configs/markets/market, because I suspect you want to copy the market elements, rather than the single markets element
• for the printXmlDocument, I've used the interesting code in this answer

I hope this helps.

---

If you don't want to create a new root element, then you may use your original XPath expression, which returns a NodeList consisting of a single node (keep in mind that your XML must have a single root element) that you can directly add to your new XML document.

See following code, where I commented lines from the code above:

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();
    }
}

This will give you the following output:

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

这篇关于使用nodeList创建XML文档的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！