分类导航

Android 发布时间：2022-04-28 发布网站：大佬教程 code.js-code.com

大佬教程收集整理的这篇文章主要介绍了在Android中使用XPath在XML文件中搜索，大佬教程大佬觉得挺不错的，现在分享给大家，也给大家做个参考。

我在 android上开发应用程序！

好吧,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.

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！！

解决方法

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 + ")");
    }
  }
}