以下是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>
以下是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为 -
Plastic Surgery
Mal-nutrition
但我也获得了下一个Subject
代码的输出,该代码的属性值为 IT ...... ???
查询中缺少什么...?
答案 0 :(得分:3)
您的查询应该是
String qry = "//Seminars/Seminar/Subject[contains(@name,'Medical')]/Topic/@main";
您要求在Seminar
中Subject
中包含@name
和Topics
的所有主题。新查询要求@name中包含“Medical”的Subject
中的所有{{1}}。微妙但重要的区别!
答案 1 :(得分:2)
您的XPath将为所有至少有一个名为 Medical 的主题的研讨会提供所有主题/主题[@main]。
你可能想要
String qry = "//Seminars/Seminar/Subject[contains(@name,'Medical')]/Topic/@main";
或
String qry = "//Seminars/Seminar/Subject[@name='Medical']/Topic/@main";