Question

我从Android设备连接到服务器并查询数据库并在屏幕上显示结果。但是我想在ListView中显示我的结果。如何让我的代码显示在ListView中？目前它只是解析数据并在屏幕上显示。如何为ContactName的每个值赋予参考值？以下是我的代码：

   public class DbConnectActivity extends Activity {
/** Called when the activity is first created. */

   TextView txt;
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    // Create a crude view - this should really be set via the layout resources 
    // but since its an example saves declaring them in the XML. 
    LinearLayout rootLayout = new LinearLayout(getApplicationContext()); 
    txt = new TextView(getApplicationContext()); 
    rootLayout.addView(txt); 
    setContentView(rootLayout); 

    // Set the text and call the connect function. 
    txt.setText("Connecting...");
  //call the method  to run the data retreival
    txt.setText(getServerData(KEY_121));



}
public static final String KEY_121 = "http://10.0.2.2/dbconnect.php"; //i use my real ip here



private String getServerData(String returnString) {

   InputStream is = null;
   String result = "";


    //the year data to send
   // ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
    //nameValuePairs.add(new BasicNameValuePair("EngID","1"));

    //http post
    try{
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost(KEY_121);
            //httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            HttpResponse response = httpclient.execute(httppost);
            HttpEntity entity = response.getEntity();
            is = entity.getContent();

    }catch(Exception e){
            Log.e("log_tag", "Error in http connection "+e.toString());
    }

    //convert response to string
    try{

            BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
            StringBuffer sb = new StringBuffer();
            String line = "";
            while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
            }
            is.close();
            result=sb.toString();
            //return result;


    }catch(Exception e){
            Log.e("log_tag", "Error converting result "+e.toString());
    }
    //parse json data
    try{
            JSONArray jArray = new JSONArray(result);
            for(int i=0;i<jArray.length();i++){
                    JSONObject json_data = jArray.getJSONObject(i);
                    Log.i("log_tag","Name: "+json_data.getString("ContactName")
                           );
                    //Get an output to the screen
                    returnString += "\n\t" + jArray.getJSONObject(i);

            }
    }catch(JSONException e){
            Log.e("log_tag", "Error parsing data "+e.toString());
    }

    return returnString;
}   

}

Answer 1

public class JsonExampleActivity extends ListActivity {
    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        //setContentView(R.layout.main);
        setListAdapter(new ArrayAdapter(this, android.R.layout.simple_list_item_1, this.fetchTwitterPublicTimeline()));
    }

    public ArrayList<String> fetchTwitterPublicTimeline()
    {
        ArrayList<String> listItems = new ArrayList<String>();

        try {
            URL twitter = new URL("your url here");

            URLConnection tc = twitter.openConnection();
            BufferedReader in = new BufferedReader(new InputStreamReader(
                    tc.getInputStream()));

            String line;
            while ((line = in.readLine()) != null) {
                  JSONObject ja = new JSONObject(line);
              /*
                //for (int i = 0; i < ja.length(); i++) {
                    JSONObject jo = ja.getJSONObject("_api_error");
                    listItems.add(jo.getString("name"));*/

                    JSONArray jobj=ja.getJSONArray("artists");
                    //JSONArray ja = new JSONArray(line);
                for (int i = 0; i < jobj.length(); i++) {
                    JSONObject jo = jobj.getJSONObject(i);
                    listItems.add(jo.getString("artist_name"));
                }
               // }
            }
        } catch (MalformedURLException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (JSONException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        return listItems;
    }
}

Answer 2

我要做的最简单的事情是将JSONArray放在某个ArrayList中并在arrayadapter中使用它

Answer 3

您可以将所有json对象添加到列表中，然后从那里加载列表，但是您已准备好返回单个字符串的方法。你可以使用以下内容：

使用

String[] tokens=returnString.split("\n\r");
ArrayAdapter adapter=new ArrayAdapter(this, android.R.layout.simple_list_item_1, tokens);
listView.setAdapter(adapter);

Answer 4

了解自定义适配器的基础知识然后......

每当你想用ListView中的视图进行处理时需要创建一个可以处理逻辑的自定义适配器实现并在必要时将该信息传递给视图。

Answer 5

我确保您在后台线程中提出请求（例如异步任务或意向服务），否则如果需要很长时间，您的应用可能会崩溃。

关于列表活动

http://www.vogella.de/articles/AndroidListView/article.html

本教程将向您展示如何创建自定义模型并使用自定义xml将其填充到列表视图中。

将JSON输出解析为ListView

5 个答案: