java - 在 XSLT 中得到错误的输出？

Question

下面是 XML 文件 -

<Seminars>
  <Seminar>
    <Venue P="ABC" dt="20111223"/>
    <Subject name="Finance">
      <Topic main="Bonds"/>
      <Topic main="Stocks" sub="Technical Analysis"/>
    </Subject>       
  </Seminar>    
  <Seminar>
    <Venue P="ABC" dt="20111225"/>
    <Subject name="Yoga">
      <Topic main="Benefits"/>
    </Subject>
    <Subject name="Meditation">
      <Topic main="Benefits"/>
    </Subject>
  </Seminar>
  <Seminar>
    <Venue P="PQR" dt="20111220"/>       
    <Subject name="IT">
       <Topic main="Java" sub="Swing"/>
       <Topic main="Java" sub="NIO"/>
    </Subject>       
  </Seminar>
  <Seminar>
    <Venue P="ABC" dt="20111224"/>
    <Subject name="Medical">
       <Topic main="Plastic Surgery"/>
       <Topic main="Mal-nutrition"/>
    </Subject>
    <Subject name="IT">
       <Topic main="Java" sub="Collections"/>
       <Topic main="Web Technologies"/>
    </Subject>      
  </Seminar>
  <Seminar>     
    <Venue P="XYZ" dt="20111226"/>
    <Subject name="IT">
      <Topic main="DotNET - I"/>
      <Topic main="DotNET - II"/>
      <Topic main="XML" sub="Security"/>
    </Subject>       
  </Seminar>
  <Seminar>
     <Venue P="XYZ" dt="20111227"/>
     <Subject name="IT">
        <Topic main="Oracle"/>
        <Topic main="Oracle" sub="JDeveloper"/>
     </Subject>       
  </Seminar>
</Seminars>

下面是Java程序-

import org.w3c.dom.*;
import javax.xml.xpath.*;
import javax.xml.parsers.*;
import java.io.IOException;
import org.xml.sax.SAXException;

public class Seminar
{
   public static void main(String[] args) throws ParserConfigurationException, SAXException, IOException, XPathExpressionException 
   {
      DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
      domFactory.setNamespaceAware(true); 
      DocumentBuilder builder = domFactory.newDocumentBuilder();
      Document doc = builder.parse("seminar.xml");
      XPath xpath = XPathFactory.newInstance().newXPath();

      String qry = "//Seminars/Seminar[contains(Subject/@name,'Medical')]/Subject/Topic/@main";
      XPathExpression expr = xpath.compile(qry);
      Object result = expr.evaluate(doc, XPathConstants.NODESET);
      NodeList nodes = (NodeList) result;

      for (int i = 0; i < nodes.getLength(); i++) 
      {
           System.out.println(nodes.item(i).getNodeValue());
      }
   }
}

使用上面的程序我期望输出为 -

Plastic Surgery
Mal-nutrition

但我也得到了下一个 Subject 标签的输出，它的属性值为 IT ...???

我在查询中遗漏了什么...？

Answer 1

你的查询应该是

  String qry = "//Seminars/Seminar/Subject[contains(@name,'Medical')]/Topic/@main";

您要求的是 Seminar 中的所有主题，其中包含 @name 中带有“Medical”的 Subject。新查询要求在@name 中包含“Medical”的Subject 中的所有Topics。细微但重要的区别!