在Android中使用XPathsearchXML文件

我正在开发Android上的应用程序!

那么我现在有一点冲突,我想执行一个XPath查询,但我没有到达解决这个问题。

在这里输入图像说明

这是我使用的XML文件的一个例子:

<?xml version="1.0"?> <catalog> <book id="bk101"> <author>Gambardella, Matthew</author> <title>XML Developer's Guide</title> <genre>Computer</genre> <price>44.95</price> <publish_date>2000-10-01</publish_date> <description>An in-depth look at creating applications with XML.</description> </book> <book id="bk102"> <author>Ralls, Kim</author> <title>Midnight Rain</title> <genre>Fantasy</genre> <price>5.95</price> <publish_date>2000-12-16</publish_date> <description>A former architect battles corporate zombies, an evil sorceress.</description> </book> <book id="bk103"> <author>Corets, Eva</author> <title>Maeve Ascendant</title> <genre>Fantasy</genre> <price>5.95</price> <publish_date>2000-11-17</publish_date> <description>After the collapse of a nanotechnology society in England.</description> </book> </catalog> 

我能怎么做??

提前致谢!!

Solutions Collecting From Web of "在Android中使用XPathsearchXML文件"

看看这个例子:

 import java.io.FileReader; import javax.xml.xpath.XPath; import javax.xml.xpath.XPathConstants; import javax.xml.xpath.XPathFactory; import org.w3c.dom.Element; import org.w3c.dom.NodeList; import org.xml.sax.InputSource; public class GuestList { public static void main(String[] args) throws Exception { XPathFactory factory = XPathFactory.newInstance(); XPath xPath = factory.newXPath(); NodeList shows = (NodeList) xPath.evaluate("/schedule/show", new InputSource(new FileReader( "tds.xml")), XPathConstants.NODESET); for (int i = 0; i < shows.getLength(); i++) { Element show = (Element) shows.item(i); String guestName = xPath.evaluate("guest/name", show); String guestCredit = xPath.evaluate("guest/credit", show); System.out.println(show.getAttribute("weekday") + ", " + show.getAttribute("date") + " - " + guestName + " (" + guestCredit + ")"); } } } 

其余的例子在这里: http : //jexp.ru/index.php/Java_Tutorial/XML/XPath