在Android中解析JSONArray

Question

我是json解析的新手，很忙。我必须解析以下内容： -

[ 
    { 
    "firstname": abc, 
    "lastname": xyp, 
    "designation" : executive, 
    "user": { 
     "username": "xypabc", 
     "userid": 4003, 
     }, 
    }, 

    { 
    "firstname": pqr, 
    "lastname": vbn, 
    "designation" : security, 
    "user": { 
     "username": "vbnpqr", 
     "userid": 11231, 
     }, 
    },  


    { 
    "firstname": ghk, 
    "lastname": lkj, 
    "designation" : manager, 
    "user": { 
     "username": "lkjghk", 
     "userid": 774, 
     }, 
    } 
] 

我需要从上面获取“login”和“userid”。下面是我写的代码： -

try { 
    JSONArray jsonObj = new JSONArray(response); 
    for(int i=0 ; i<jsonObj.length(); i++) 
    {          
     JSONObject json_Data = jsonObj.getJSONObject(i); 
     String userName = json_Data.getString("username"); 
     String userId = json_Data.getString("userid"); 
     Log.d("Factors","UserName :- "+userName+" ID :- "+userId); 
    } 
    }catch (JSONException e) { 
     Log.d("Failure","Dude I have failed"); 
    }. 

问题是我的代码以异常结束。 请帮忙!!!

Answer 1

username和userid是user内的JSONObject分析用户的JSONObject，然后获取用户名和用户ID的字符串。

DO这样得到username和userid

for(int i=0 ; i<jsonObj.length(); i++) 
    {          
     JSONObject json_Data = jsonObj.getJSONObject(i); 
     String userName = json_Data.getJSONObject("user").getString("username"); 
     String userId = json_Data.getJSONObject("user").getString("userid"); 
     Log.d("Factors","UserName :- "+userName+" ID :- "+userId); 
    } 

Answer 2

解析：“用户”

try { 
     JSONArray jsonObj = new JSONArray(response); 
     for(int i=0 ; i<jsonObj.length(); i++) 
     {          
      JSONObject json_Data = jsonObj.getJSONObject(i); 
      JSONObject user = json_Data.getJSONObject("user"); 
      String userName = user.getString("username"); 
      String userId = user.getString("userid"); 
      Log.d("Factors","UserName :- "+userName+" ID :- "+userId); 
     } 
     }catch (JSONException e) { 
      Log.d("Failure","Dude I have failed"); 
     } 

Answer 3

JSONArray jresult = new JSONArray(response); 

jresult = json.getJSONArray("user"); 
for (int i = 0; i < jresult .length(); i++) 
{ 
JSONObject obj = jresult .getJSONObject(i); 
String username=obj.getString("username"); 
int userid=obj.getInt("userid"); 
} 

Answer 4

if(result != null) 
      { 
       try 
       { 
        JSONObject jobj = result.getJSONObject("result"); 

        String status = jobj.getString("status"); 

        if(status.equals("true")) 
        { 
         JSONArray array = jobj.getJSONArray("user"); 

         for(int x = 0; x < array.length(); x++) 
         { 
          HashMap<String, String> map = new HashMap<String, String>(); 

          map.put("username", array.getJSONObject(x).getString("username")); 

          map.put("userid", array.getJSONObject(x).getString("userid")); 



          list.add(map); 
         } 

         CalendarAdapter adapter = new CalendarAdapter(Calendar.this, list); 

         list_of_calendar.setAdapter(adapter); 
        } 
       } 
       catch (Exception e) 
       { 
        e.printStackTrace(); 
       } 

Answer 5

试试这个代码：:)

JSONArray jsonObj = new JSONArray(response); 
       for(int i=0 ; i<jsonObj.length(); i++) 
       {          
        JSONObject json_Data = jsonObj.getJSONObject(i); 
        String firstname = json_Data.getString("firstname"); 
        String lastname = json_Data.getString("lastname"); 
        String designation = json_Data.getString("designation"); 

        JSONArray jsons1 = json_Data.getJSONArray("user"); 
        for (int j = 0; j < jsons1.length(); j++) { 

         JSONObject jsonss = jsons1.getJSONObject(j); 
         String username = jsonss.getString("username"); 
          String userid = jsonss.getString("userid"); 
        } 

       } 
       }catch (JSONException e) { 
        Log.d("Failure","Dude I have failed"); 
       }