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我的列表视图中有超过 10 行，我正在使用 JSON 从 Web URL 获取记录，但现在我想为每一行调用特定的意图，例如：-对于第一行，我想调用 ( this)，对于第二行，( http://domainname.com/s/b）等等，所以请建议我需要在我的 ListActivity 类中添加哪些代码以及在哪里调用......

public class MainActivity extends Activity {

ListView mListView;

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

    // URL to the JSON data 
    String strUrl = "http://domainname.in/first.php/count/";

    // Creating a new non-ui thread task to download json data 
    DownloadTask downloadTask = new DownloadTask();

    // Starting the download process
    downloadTask.execute(strUrl);

    // Getting a reference to ListView of activity_main
    mListView = (ListView) findViewById(R.id.lv_count);
 }
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2 回答 2

1

您可以通过编写自己的自定义适配器来实现这一点。如下所示：

  class CustomAdapter extends ArrayAdapter {

    HashMap<String, String> map;
    ArrayList mList;
    public CustomAdapter(Context context, int resource,
            int textViewResourceId, HashMap<String, String> map) {
        super(context, resource, textViewResourceId);
        this.map = map;
        mList = new ArrayList<String>(map.keySet());
    }

    @Override
    public int getCount() {
        return mList.size();
    }

    @Override
    public Object getItem(int position) {
        return mList.get(position);
    }

    @Override
    public View getView(int position, View convertView, ViewGroup parent) {
        // do your stuff with layouts.
        String item = (String)mList.get(position);
        // either you can set the url as tag to view so that the same url will be
        //retrieved when an item is clicked or declare the map as global variable.
        String url = map.get(item);
        convertView.setTag(url);
        return convertView;
    }
}

/// 在你的活动中

@Override
public void onItemClick(AdapterView<?> adapter, View view,
        int position, long id) {
    // you can get the url from the tag
    String url = view.getTag().toString();
    // or else declare the HashMap as global variable which will be accessible through out the class.
    String url = new ArrayList<String>(map.keySet()).get(position);
    // Now launch web with this url
}
于 2012-09-24T06:01:09.673 回答
0

您需要ListView使用为每个项目指定的 URL 附加数据，并且您可以处理 ListViewonItemClick()从选定项目的基础获取 ListView 项目获取 URL 并在 WebView 中打开链接。

于 2012-09-24T05:19:31.423 回答