Question

我正在编写HTTP服务器，我将在HTTP PUT中接收XPATH，并在请求正文中接收数据。

我需要将XPATH表达式的结果替换为XML文档中HTTP请求数据中的数据

例如

XML文档是

<presence>
<tuple id="x8eg92n">
<note> i am reading email 3 times a day </note>
</tuple>
</presence>

HTTP请求例如是

  PUT /pidf-manipulation/users/sip:someone@example.com/index/
  ~~/presence/tuple%5b@id='x8eg92n'%5d/note HTTP/1.1
  If-Match: "xyz"
  Host: xcap.example.com
  Content-Type: application/xcap-el+xml
  ...

  <note>I'm reading mails on Tuesdays and Fridays</note>

上面的内容应该用PUT请求替换XML中的note元素。客户端可以通过这种方式发送任何XPATH并替换XML文档的内容。

请帮助您了解如何在Java代码中完成此操作。

Answer 1

基本上你需要做的是：

加载要更改的xml;
在替换中加载要使用的片段;
在[1]中加载的文档中采用[2]中加载的节点;
使用XPath查找要替换的节点;
从[4]中找到的父节点中获取父节点，从父节点中删除[4]中找到的节点，然后添加您采用的节点。

你可以使用类似的东西：

import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.StringWriter;

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerException;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathExpression;
import javax.xml.xpath.XPathExpressionException;
import javax.xml.xpath.XPathFactory;

import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.xml.sax.SAXException;

public class XmlXPathReplace {

    public static void main(final String[] args) throws SAXException, IOException, ParserConfigurationException,
            XPathExpressionException, TransformerException {
        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
        factory.setNamespaceAware(true); // never forget this!
        DocumentBuilder builder = factory.newDocumentBuilder();

        // step 1.
        Document doc = builder.parse(new ByteArrayInputStream(( //
                "<?xml version=\"1.0\"?>" + //
                        "<people>" + //
                        "<person><name>First Person Name</name></person>" + //
                        "<person><name>Second Person Name</name></person>" + //
                        "</people>" //
                ).getBytes()));

        // step 2
        String fragment = "<name>Changed Name</name>";
        Document fragmentDoc = builder.parse(new ByteArrayInputStream(fragment.getBytes()));

        // step 3
        Node injectedNode = doc.adoptNode(fragmentDoc.getFirstChild());

        // step 4
        XPath xPath = XPathFactory.newInstance().newXPath();
        XPathExpression expr = xPath.compile("//people/person[2]/name");
        System.out.println();
        Element nodeFound = (Element) expr.evaluate(doc, XPathConstants.NODE);

        // step 5
        Node parentNode = nodeFound.getParentNode();
        parentNode.removeChild(nodeFound);
        parentNode.appendChild(injectedNode);

        DOMSource domSource = new DOMSource(doc);
        Transformer transformer = TransformerFactory.newInstance().newTransformer();
        StreamResult result = new StreamResult(new StringWriter());
        transformer.transform(domSource, result);
        System.out.println(result.getWriter().toString());
    }

}

Answer 2

您还可以使用以下命令，而不是删除旧节点并添加新节点： parentNode.replaceChild(injectedNode, nodeFound)。使用此功能，您可以保持节点顺利

Answer 3

Node importedTemplateChildNode = targetDoc.importNode(templateChildNode, true);   
// Importing template node to the target document(this solves wrong_DOCUMENT_ERR:)

targetParentNode.replaceChild(importedTemplateChildNode, targetChildnode);       
// Replace target child node with the template node

Transformer tranFac =TransformerFactory.newInstance().newTransformer();
tranFac.transform(new DOMSource(targetDoc), new StreamResult(new FileWriter(targetXmlFile)));

使用Java中的String替换XML文档中的元素

3 个答案: