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我是 android 开发新手，我想使用 Dom Parser 解析一个特定的 xml 文件。xml 结构是这样的。

<Images>
   <image link="a.tgcdn.net/images/products/zoom/e554_android_plush_robot.jpg"/>
   <image link="a.tgcdn.net/images/products/zoom/e554_android_plush_robot.jpg"/>
   <image link="http://cdn.androidpolice.com/wp-content/themes/ap1/images/android1.png"/>
   <image link="http://cdn.androidpolice.com/wp-content/themes/ap1/images/android1.png"/>
   <image link="http://cdn.androidpolice.com/wp-content/themes/ap1/images/android1.png"/>
</Images>

我想将这些 Web 链接添加到数组列表中。建议我解决它..甚至与此相关的任何链接或教程都会有所帮助。谢谢。

这是完整的代码..

public class MainActivity extends Activity {


    String xmlurl       = "--url of xml---";


    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        ArrayList<String> mImageLink = new ArrayList<String>();
        try {
            URL url = new URL(xmlurl);
            XMLParser   parser  =   new XMLParser();
            String      xml     = parser.getXMLfromUrl(url);
            Document    doc     = parser.getDomElement(xml);
            DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
            DocumentBuilder db = dbf.newDocumentBuilder();
            Document doc1 = db.parse(new InputSource(url.openStream()));
            doc1.getDocumentElement().normalize();

            NodeList nodeList = doc1.getElementsByTagName("photo");

            for (int i = 0; i < nodeList.getLength(); i++) {
                Element websiteElement = (Element) nodeList.item(i);
                nodeList = websiteElement.getChildNodes();              
                mImageLink.add(websiteElement.getAttribute("link"));

            }
        } catch (Exception e) {
            System.out.println("XML Pasing Excpetion = " + e);
        }

        for(int i=0;i<mImageLink.size();i++){
            Log.d("Photo link --- " + i,mImageLink.get(i));
        }

    }

    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        getMenuInflater().inflate(R.menu.activity_main, menu);
        return true;
    }
}
4

2 回答 2

0

请使用以下代码使用 DOM Parser 解析上述 XML 文件。

public class MainActivity extends Activity {

    ArrayList<String> mImageLink;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        try {
            mImageLink = new ArrayList<String>();

            InputStream is = getResources().openRawResource(R.raw.temp);
            DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
            DocumentBuilder db = dbf.newDocumentBuilder();
            Document doc = db.parse(new InputSource(is));
            doc.getDocumentElement().normalize();

            NodeList nodeList = doc.getElementsByTagName("image");

            for (int i = 0; i < nodeList.getLength(); i++) {

                Node node = nodeList.item(i);

                Element fstElmnt = (Element) node;

                mImageLink.add(fstElmnt.getAttribute("link"));

            }
        } catch (Exception e) {
            System.out.println("XML Pasing Excpetion = " + e);
        }
    }
}

有关更多信息，请参见下面的链接。

使用 DOM 解析器进行 XML 解析

于 2012-12-04T09:00:46.543 回答
0

使用dom4j库，它是 SAXReader。 

    InputStream is = FileUtils.class.getResourceAsStream("file.xml");

    SAXReader reader = new SAXReader();
    org.dom4j.Document doc = reader.read(is);
    is.close();
    Element content = doc.getRootElement();  //this will return the root element in your xml file
    List<Element> methodEls = content.elements("element"); // this will retun List of all Elements with name "element"

对于 DOM 解析，您可以使用同一个库的DOMReader

于 2012-12-04T08:06:46.073 回答