我怎樣才能在android中獲得JSON響應？

Question

我想爲我的應用程序獲取json數據。

這裏是CallAPI.java類來獲得響應

public class CallAPI extends AsyncTask<String,String,String> { 

    String jsonStr=""; 
    private static final String url="https://api.litzscore.com/rest/v2/recent_matches/"; 

    @Override 
    protected String doInBackground(String... params) { 
     List<NameValuePair> param=new ArrayList<>(); 
     param.add(new BasicNameValuePair("access_token",""));//here i am added my access token 
     ServiceHandler sh=new ServiceHandler(); 
     jsonStr=sh.makeServiceCall(url,ServiceHandler.GET,param); 

     try{ 
      JSONObject obj=new JSONObject(jsonStr);//I got error here 

      Log.e("response",""+obj.toString()); 
     } 
     catch (JSONException e){ 
      e.printStackTrace(); 
     } 
     return null; 
    } 
} 

這裏是我的JSON ServiceHandler.java

public class ServiceHandler { 
    static String response=null; 
    public final static int GET = 1; 
    public final static int POST = 2; 

    public ServiceHandler() { 

    } 

    public String makeServiceCall(String url, int method) { 
     return this.makeServiceCall(url, method, null); 
    } 
    public String makeServiceCall(String url, int method, 
            List<NameValuePair> params) { 
     try { 
      DefaultHttpClient httpClient = new DefaultHttpClient(); 
      HttpEntity httpEntity = null; 
      HttpResponse httpResponse = null; 

      if (method == POST) { 
       HttpPost httpPost = new HttpPost(url); 
       if (params != null) { 
        httpPost.setEntity(new UrlEncodedFormEntity(params)); 
       } 

       httpResponse = httpClient.execute(httpPost); 

      } else if (method == GET) { 
       if (params != null) { 
        String paramString = URLEncodedUtils 
          .format(params, "utf-8"); 
        url += "?" + paramString; 
       } 
       Log.e("URL",""+url); 
       HttpGet httpGet = new HttpGet(url); 

       httpResponse = httpClient.execute(httpGet); 
      } 
      httpEntity = httpResponse.getEntity(); 
      response = EntityUtils.toString(httpEntity); 

     } catch (UnsupportedEncodingException e) { 
      e.printStackTrace(); 
     } catch (ClientProtocolException e) { 
      e.printStackTrace(); 
     } catch (IOException e) { 
      e.printStackTrace(); 
     } 
     Log.e("RESPONSE",response); 
     return response; 

    } 

} 

這裏是我的錯誤

org.json.JSONException: Value ���nҩV�͗Qo�0ǿ��ץ`�o���j+� of type java.lang.String cannot be converted to JSONObject 

如果我叫從本地服務器，如文件wammp，xampp然後它工作正常。 請幫我解決這個問題。

Answer 1

嘗試轉換這樣

public class ServiceHandler { 
static String response=null; 
InputStream is = null; 
public final static int GET = 1; 
public final static int POST = 2; 

public ServiceHandler() { 

} 

public String makeServiceCall(String url, int method) { 
    return this.makeServiceCall(url, method, null); 
} 
public String makeServiceCall(String url, int method, 
           List<NameValuePair> params) { 
    try { 
     DefaultHttpClient httpClient = new DefaultHttpClient(); 
     HttpEntity httpEntity = null; 
     HttpResponse httpResponse = null; 

     if (method == POST) { 
      HttpPost httpPost = new HttpPost(url); 
      if (params != null) { 
       httpPost.setEntity(new UrlEncodedFormEntity(params)); 
      } 

      httpResponse = httpClient.execute(httpPost); 

     } else if (method == GET) { 
      if (params != null) { 
       String paramString = URLEncodedUtils 
         .format(params, "utf-8"); 
       url += "?" + paramString; 
      } 
      Log.e("URL",""+url); 
      HttpGet httpGet = new HttpGet(url); 

      httpResponse = httpClient.execute(httpGet); 
     } 
     httpEntity = httpResponse.getEntity(); 
     is = httpEntity.getContent(); 

     //convert response to string 
     BufferedReader reader = new BufferedReader(new InputStreamReader(is, "iso-8859-1"), 8); 
     StringBuilder sb = new StringBuilder(); 
     String line = null; 
     while ((line = reader.readLine()) != null) { 
      sb.append(line + "\n"); 

     } 
     is.close(); 
     response = sb.toString(); 

    } catch (UnsupportedEncodingException e) { 
     e.printStackTrace(); 
    } catch (ClientProtocolException e) { 
     e.printStackTrace(); 
    } catch (IOException e) { 
     e.printStackTrace(); 
    } 
    Log.e("RESPONSE",response); 
    return response; 

} 

}