Question

我正在尝试使用JSON和HTTP客户端从我的数据库中的表中检索数据。我的代码似乎正在工作，但它没有输出任何东西，但如果我在查询中故意更改错误吐司输出，所以我知道代码是正确的没有错误输出。问题在于弄清楚如何将JSON输出放到我的视图屏幕上。

以下是完整的HTTP代码：

public void LoadRemote()
{

     ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
    //http post
    try{
         HttpClient httpclient = new DefaultHttpClient();
         HttpPost httppost = new HttpPost("http://deucalion0.co.uk/getscores.php");
         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);
        sb = new StringBuilder();
        sb.append(reader.readLine() + "\n");

        String line="0";
        while ((line = reader.readLine()) != null) {
                       sb.append(line + "\n");
         }
         is.close();
         result=sb.toString();
         }catch(Exception e){
               Log.e("log_tag", "Error converting result "+e.toString());
         }
 //paring data
 int score;
 String username;
 try{
       jArray = new JSONArray(result);
       JSONObject json_data=null;
       for(int i=0;i<jArray.length();i++){
              json_data = jArray.getJSONObject(i);
              score=json_data.getInt("score");
              username=json_data.getString("username");
          }
       }
       catch(JSONException e1){
          Toast.makeText(getBaseContext(), "No City Found" ,Toast.LENGTH_LONG).show();
       } catch (ParseException e1) {
            e1.printStackTrace();
    }
 }

这是我的SQL查询：

$sql=mysql_query("select score, username from highscores ");
while($row=mysql_fetch_assoc($sql))
$output[]=$row;
print(json_encode($output));

您可以看到此工作here

我很感激有关这个问题的任何建议。

感谢。

由于并发症，我已经重新开始，这里是活动类，现在是：

public class Viewscores extends Activity{

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

    scores = (Button) findViewById(R.id.bViewscores);

    scores.setOnClickListener(new View.OnClickListener(){

        public void onClick(View v) {

            LoadRemote();
        }

    });

}


  public void LoadRemote()
    {

   String result = "";
   ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
     InputStream is = null;
     try{
         HttpClient httpclient = new DefaultHttpClient();
         HttpPost httppost = new                HttpPost("http://deucalion0.co.uk/getscores.php");
         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());
        }

BufferedReader myReader = new BufferedReader(new InputStreamReader(is));



    }


     }

Answer 1

我可能错了，但是如果你创建一个像这样的包装器对象并不容易：

public class Response {
        public int score;
        public String username;
    }

然后用gson反序列化json很容易：

ArrayList<Response> responses = new ArrayList<Response>();
Gson gson = new Gson();
Type collectionType = new TypeToken<List<Response>>(){}.getType();
responses = gson.fromJson(response, collectionType);

其中response是HTTP请求中的String。有关gson的更多信息： http://code.google.com/p/google-gson/

Answer 2

我无法完全理解你的意思是“将JSON输出放到我的视图屏幕上”..因为你没有提到你正在使用的视图..所以假设你使用listview这个......这就是什么我认为可以做到。

try{
   jArray = new JSONArray(result);
   int k=jArray.length();
   String s[]=new String[k];
     for(int u=0;u<k;u++){
         json_data = jArray.getJSONObject(i);
          s[u]=json_data.getInt("score")+"  "+json_data.getString("username");
        }
           setContentView(R.layout.fine);
           ListView lv = (ListView)findViewById(R.id.list);
           ArrayAdapter<String> adapter = new ArrayAdapter<String>(this,
                R.layout.list,R.id.textView1,s);
           lv.setAdapter(adapter);

Answer 3

好的..这是整个代码...

public class Viewscores extends Activity{

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

scores = (Button) findViewById(R.id.bViewscores);

scores.setOnClickListener(new View.OnClickListener(){

    public void onClick(View v) {

        Intent i=new Intent(this,lst.class);
startActivity(i);
    }

});

}

这里是“lst”类

public class lst extends ListActivity{
JSONArray jArray; 
String result = null; 
InputStream is = null; 
StringBuilder sb=null;  
String s[];

@Override 
public void onCreate(Bundle savedInstanceState) { 
super.onCreate(savedInstanceState); 

try{ 
URL u=new URL("http://deucalion0.co.uk/getscores.php");
URLConnection tc=u.openConnection();
BufferedReader reader = new BufferedReader(new InputStreamReader(tc.getInputStream())); 


String line; 

while ((line = reader.readLine()) != null) { 
jArray =new JSONArray(line);
s=new String[jArray.length()];
for(int m=0;m<jArray.length();m++){
    JSONObject json_data=(JSONObject)jArray.get(m);

    s[m]=json_data.getInt("score")+" "+json_data.getString("username"); 
}
} 

ArrayAdapter<String> adapter = new ArrayAdapter<String>(this, 
android.R.layout.simple_list_item_1, s); 
setListAdapter(adapter); 
} 

catch(JSONException e1){ 
} catch (ParseException e1) { 
e1.printStackTrace(); 
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} 
}                   |
}                                        ---

这个肯定有用......试试这个......：）

在检索SQL查询时如何使用android中的JSON输出到视图

3 个答案: