Question

这是我的要求，

我从服务器获得响应作为soap对象，我将其转换为字符串格式，现在我想提取xml数据，

如何提取嵌套标签？

我可以提取所有数据还是需要解析该XML响应？有没有办法提取XML嵌套标签并将其解析为ArrayList＆lt;＆gt;哪个帖子是一个班级？

<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>
..

如果可能的话怎么办？

先谢谢。

我看到可以使用它以及如何使用

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;
    }
}

Answer 1

您可以使用JAXB来编组/解组SOAP数据包。我假设你有一个你拥有的SOAP数据包的模式。看看这个 - JAXB。

Answer 2

您可以像普通XML一样解析SOAP对象（SAX等）。查看本教程： http://www.mkyong.com/java/how-to-read-xml-file-in-java-sax-parser/

XPath： Soap response message parsing using XPath

并且：Java library to form and parse soap messages

并且：http://www.coderanch.com/t/520979/XML/parse-SOAP-Xml-java-code

如何从Soap对象中提取XML嵌套标签？

2 个答案: