在Android中使用XPath搜索XML文件

我正在開發android上的應用程序！在Android中使用XPath搜索XML文件

那麼我現在有一點衝突，我想執行一個XPath查詢，但我沒有到達解決這個問題。

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這

XML filethat我用一個例子：

<?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>

我該怎麼辦？

在此先感謝！

來源

2011-04-01 alibenmessaoud

請看下面的例子：

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 + ")"); 
    } 
    } 
}

的示例都在這裏：http://jexp.ru/index.php/Java_Tutorial/XML/XPath

來源

2011-04-14 14:03:09 alibenmessaoud

+1只是爲了被別人誰記得，顯示所導入的命名空間！ – SteveCav 2012-02-01 01:48:05

如果我在線託管XML，我可以解析它，我應該讀取它全部 – AMH 2012-07-29 00:13:42

要讀取外部XML文件，可以使用URL類使用SAX打開流。用此行代替NodeList實例化NodeList shows =（NodeList）xPath.evaluate（「/ schedule/show」，new InputSource（new URL（「http://www.exple.com/file.xml").openStream （）），XPathConstants.NODESET）;' – alibenmessaoud 2012-07-29 10:47:05

在Android中使用XPath搜索XML文件

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