android编程布局显示的问题

Question

运行结果如图所示，但是运行结果不是我想要的，获取的新闻不能显示在listview上，想要的结果类似于那个点菜那样，如图所示
public class MainActivity extends Activity {

Document doc;

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

    findViewById(R.id.button1).setOnClickListener(new OnClickListener() {
        @Override
        public void onClick(View v) {
            load();
        }
    });
}

protected void load() {

    try {
        doc = Jsoup.connect("http://www.gdst.cc/News.aspx?id=1").get();
    } catch (MalformedURLException e1) {
        e1.printStackTrace();
    } catch (IOException e1) {
        e1.printStackTrace();
    }

    List<Map<String, String>> list = new ArrayList<Map<String, String>>();
    Elements es = doc.getElementsByClass("single_page_list");
    for (Element e : es) {
        Map<String, String> map = new HashMap<String, String>();
        map.put("title", e.getElementsByTag("li").text());
        map.put("href", "http://www.gdst.cc/"
                + e.getElementsByTag("li").first().getElementsByTag("a").first().attr("href"));
        list.add(map);
    }

    ListView listView = (ListView) findViewById(R.id.listView1);
    SimpleAdapter Adapter = new SimpleAdapter(this, list, android.R.layout.simple_list_item_2,
            new String[] { "title","href" }, new int[] {
            android.R.id.text1,android.R.id.text2
    });
    listView.setAdapter(Adapter);


}

/**
 * @param urlString
 * @return
 */
public String getHtmlString(String urlString) {
    try {
        URL url = null;
        url = new URL(urlString);

        URLConnection ucon = null;
        ucon = url.openConnection();

        InputStream instr = null;
        instr = ucon.getInputStream();

        BufferedInputStream bis = new BufferedInputStream(instr);

        ByteArrayBuffer baf = new ByteArrayBuffer(500);
        int current = 0;
        while ((current = bis.read()) != -1) {
            baf.append((byte) current);
        }
        return EncodingUtils.getString(baf.toByteArray(), "gbk");
    } catch (Exception e) {
        return "";
    }
}

}

Answer 1

这个要具体看下是什么问题的