在 Android 中使用 XPath 在 XML 文件中搜索

Question

我正在android上开发一个应用程序！

好吧，我现在有点冲突，我想执行一个 XPath 查询，但我没有来解决这个问题。

这是我使用的 XML 文件示例：

 <?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>

我该怎么办？？

提前致谢！！

Answer 1

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