2013-07-25 44 views
5

我在使用json时遇到异常。我JSONPresr类是遵循org.json.JSONObject无法转换为JSONArray

import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStream; 
import java.io.InputStreamReader; 
import java.io.UnsupportedEncodingException; 

import org.apache.http.HttpEntity; 
import org.apache.http.HttpResponse; 
import org.apache.http.client.ClientProtocolException; 
import org.apache.http.client.methods.HttpPost; 
import org.apache.http.impl.client.DefaultHttpClient; 
import org.json.JSONException; 
import org.json.JSONObject; 

import android.os.AsyncTask; 
import android.util.Log; 

public class JSONParser { 
static InputStream is = null; 
    static JSONObject jObj = null; 
    static String json = ""; 

    // constructor 
    public JSONParser() { 

    } 

    public JSONObject getJSONFromUrl(String url) { 

     // Making HTTP request 
     try { 
      // defaultHttpClient 
      DefaultHttpClient httpClient = new DefaultHttpClient(); 
      HttpPost httpPost = new HttpPost(url); 

      HttpResponse httpResponse = httpClient.execute(httpPost); 
      HttpEntity httpEntity = httpResponse.getEntity(); 
      is = httpEntity.getContent();   

     } catch (UnsupportedEncodingException e) { 
      e.printStackTrace(); 
     } catch (ClientProtocolException e) { 
      e.printStackTrace(); 
     } catch (IOException e) { 
      e.printStackTrace(); 
     } 

     try { 
      BufferedReader reader = new BufferedReader(new InputStreamReader(
        is, "UTF-8"), 8); 
      StringBuilder sb = new StringBuilder(); 
      String line = null; 
      while ((line = reader.readLine()) != null) { 
       sb.append(line + "\n"); 
      } 
      is.close(); 
      json = sb.toString(); 
      System.out.println(" value in json sting"+json); 
     } catch (Exception e) { 
      Log.e("Buffer Error", "Error converting result " + e.toString()); 
     } 

     // try parse the string to a JSON object 
     try { 
      jObj = new JSONObject(json); 
     } catch (JSONException e) { 
      Log.e("JSON Parser", "Error parsing data " + e.toString()); 
      Log.e("JSON Parser", "Error parsing data [" + e.getMessage()+"] "+json); 
      System.out.println(" error is here"); 

     } 

     // return JSON String 
     return jObj; 

    } 

    } 

我对这些代码

  String TAG_user_detail = "user_details"; 
       String TAG_user_id = "user_id"; 
      String TAG_user_name = "user_name"; 
      String TAG_user_phone = "user_phone"; 
      String TAG_ref_id = "ref_id"; 
      JSONArray user_detail_jsonarray = null; 
      JSONParser jParser = new JSONParser(); 

      // getting JSON string from URL 
      JSONObject json = jParser.getJSONFromUrl(url); 
      try { 
       // Getting Array of Contacts 
       user_detail_jsonarray = json.getJSONArray(TAG_user_detail); 

       // looping through All Contacts 
       for(int i = 0; i < user_detail_jsonarray.length(); i++){ 
        JSONObject c = user_detail_jsonarray.getJSONObject(i); 

        // Storing each json item in variable 
        int id = c.getInt(TAG_user_id); 
        System.out.println("user id from json"+id); 

        String name_fromjson = c.getString(TAG_user_name); 
        System.out.println("user name from json"+name_fromjson); 

        int phone_no_fromjson = c.getInt(TAG_user_phone); 
        System.out.println("user phone from json"+phone_no_fromjson); 

        int ref_id_fromjson = c.getInt(TAG_ref_id); 
        System.out.println("user ref id from json"+ref_id_fromjson); 





       } 
      } catch (JSONException e) { 
       e.printStackTrace(); 
      } 

得到我的数据,但我countiniously在这一行越来越异常

user_detail_jsonarray = json.getJSONArray(TAG_user_detail); 

我赌注追踪消息是

 {"ref_id":1295,"user_name":"chand","user_phone":"9620085675","user_id":"1"} at user_details of type org.json.JSONObject cannot be converted to JSONArray 

我尝试了很多链接,但不幸的是我无法获得我的结果。请任何机构只是帮助我解决这个问题。在此先感谢所有

+2

您正在为'TAG_user_detail'获取JSONObject而不是数组 –

+0

请问您可以发布一些代码 – DJhon

+0

哪些代码?把你的完整的JSON。 –

回答

21

它清楚,你正试图JSON对象转换成JSON数组错误。那不应该。

这是读取您的JSON响应的代码。

String json = "Assuming that here is your JSON response"; 
try { 
    JSONObject parentObject = new JSONObject(json); 
    JSONObject userDetails = parentObject.getJSONObject("user_details"); 

    //And then read attributes like    
    String name = userDetails.getString("user_name"); 
    String phone = userDetails.getString("user_phone"); 
    String id = userDetails.getString("re‌​f_id"); 

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

以上代码适用于{"user_details":{"user_id":"1","user_name":"chand","user_phone":"9620085675","re‌​f_id":6386}} JSON。

+0

亲爱的Pankaj我只是通过改变退货类型来解决我的问题,而且你们帮助.... \ – DJhon

+0

好的..对不起,我...我处于疯狂状态的边缘..但是我向你投诉 – DJhon

+0

嘿...希望你将得到骑在那些.. :)所有最好的 –

3

您的jsondata是json对象格式。改变这一行

JsonObject user_detail_jsonobj = json.getJSONObject(TAG_user_detail); 
7

从异常

org.json.JSONObject cannot be converted to JSONArray 

你得到这个异常org.json.JSONObject cannot be converted to JSONArray ,因为你正试图转换JSONObject to JSONArray这是不可能的了解。

{代表JSON对象节点

[表示JSON数组节点

+0

感谢您的回复...请问您在这里发布一些代码,,,,,它将帮助完整我 – DJhon

+0

@BlueGreen通过解析JSONObject! –

+1

@BlueGreen如果你想检索用户名比使用:String name_fromjson =(String)json.get(“user_name”);不需要for循环以及jsonarray! –

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