如何从Soap对象中提取XML嵌套标记？

-4

我正在从服务器作为SOAP对象的反应，我把它转换成字符串格式，现在我想提取XML数据，

我如何可以提取嵌套的标签？

我可以提取所有数据还是需要解析XML响应？有什么办法来提取XML嵌套标签，并解析它到ArrayList <>哪里post是一个类？

<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/"> 
<soap:Body> 
    <maintag> 
    <item> 
     <name>AndroidPeople</name> 
     <website category="android">www.androidpeople.com</website> 
    </item> 
    <item> 
     <name>iPhoneAppDeveloper</name> 
     <website category="iPhone">www.iphone-app-developer.com</website> 
     </item> 
    </maintag> 
..

如果可能的话怎么样？

在此先感谢。

我看到这是有可能使用它，以及如何

private static Show getShowFromRSS(Context context, Document feed, 
     String feedUrl) { 
    try { 
     // There should be one channel in the feed, get it. 
     // Also, the cast should be okay if the XML is formatted correctly 
     NodeList item = feed.getElementsByTagName("channel"); 
     Element el = (Element)item.item(0); 

     String title; 
     NodeList titleNode = el.getElementsByTagName("title"); 
     if (titleNode == null || titleNode.getLength() < 1) 
      title = context.getString(R.string.default_title); 
     else 
      title = titleNode.item(0).getFirstChild().getNodeValue(); 

     String author; 
     NodeList authorNode = el.getElementsByTagName("author"); 
     if (authorNode == null || authorNode.getLength() < 1) 
      author = context.getString(R.string.default_author); 
     else 
      author = authorNode.item(0).getFirstChild().getNodeValue(); 

     String desc; 
     NodeList descNode = el.getElementsByTagName("comments"); 
     if (descNode == null || descNode.getLength() < 1) 
      desc = context.getString(R.string.default_comments); 
     else 
      desc = descNode.item(0).getFirstChild().getNodeValue(); 

     String imageUrl; 
     NodeList imagNode = el.getElementsByTagName("image"); 
     if(imagNode != null) { 
      Element ima = (Element)imagNode.item(0); 
      if (ima != null) { 
       NodeList urlNode = ima.getElementsByTagName("url"); 
       if(urlNode == null || urlNode.getLength() < 1) 
        imageUrl = null; 
       else 
        imageUrl = 
         urlNode.item(0).getFirstChild().getNodeValue(); 
      } else 
       imageUrl = null; 
     } else 
      imageUrl = null; 

     return new Show(title, author, feedUrl, desc, imageUrl, -1, -1); 
    } catch (Exception e) { 
     // Any parse errors and we'll log and fail 
     Log.e("NCRSS", "Error parsing RSS", e); 
     return null; 
    } 
}

来源

2013-04-05 yakusha

您需要一个XML解析器。通过这篇文章看看：http://stackoverflow.com/questions/373833/best-xml-parser-for-java – 2013-04-05 06:03:36

我更新我的问题 – yakusha 2013-04-05 06:13:41