在XSLT中获取错误的输出？

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待输出 -

Plastic Surgery 
Mal-nutrition 

但我也越来越下一个Subject标签的输出属性值为IT ... ???

我在查询中缺少什么...？

Answer 1

您的查询应该是

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

你问所有的Seminar在@name包含与“医学”一Subject的主题。新查询请求在@name中包含“Medical”的Subject中的所有Topics。微妙但重要的区别！

Answer 2

您的XPath会为您提供所有主题/主题[@main]，任何研讨会至少有一个名为医疗的主题。

你可能想

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

或

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