java - 用xpath更新xml

我只想更改jojo餐厅的等级，我尝试了此代码，但是没有用，请帮助。以下是我的xml文档和我拥有的代码。

<city>
    <beirut>
        <restaurant>
            <name>sada</name>
        </restaurant>
    </beirut>
    <jbeil>
        <restaurant>
            <name>toto</name>
            <rating>4.3/5</rating>
        </restaurant>
        <restaurant>
            <name>jojo</name>
            <rating>4.3/5</rating>
        </restaurant>
    </jbeil>
    <sour>
        <restaurant>
            <name>sada</name>
        </restaurant>
    </sour>
</city>

码：

  try {
     File inputFile = new File("src/xpath/josephXml.xml");
     DocumentBuilderFactory dbFactory
        = DocumentBuilderFactory.newInstance();
     DocumentBuilder dBuilder;

     dBuilder = dbFactory.newDocumentBuilder();

     Document doc = dBuilder.parse(inputFile);
     doc.getDocumentElement().normalize();

     XPath xPath =  XPathFactory.newInstance().newXPath();

     String expression = "/City/Jbeil/Restaurants/Restaurant[name='Feniqia']/rating";


     Element e = (Element)xPath.evaluate(expression, doc, XPathConstants.NODE);

     if (e != null){
        e.setTextContent("5/5");
     }

  } catch (ParserConfigurationException | SAXException | IOException e) {
  }

最佳答案

如果您只需要写结果（还有其他方法）：

导入java.io.File;

import org.w3c.dom.Document;
import javax.xml.transform.Result;
import javax.xml.transform.Source;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;


try
{
Transformer trans = TransformerFactory.newInstance().newTransformer();
Result out = new StreamResult(new File("result.xml"));
Source in = new DOMSource(doc);
trans.transform(in, out);
}
catch (Exception exe)
{
System.err.println(exe.getMessage());
}