AQuery JSonobject AjaxStatus错误

Question

Android代码

strUID = ((AQuery)aq.id(R.id.login_id)).getText().toString();
        strPW = ((AQuery)aq.id(R.id.login_password)).getText().toString();

        HashMap localHashMap = new HashMap();
        localHashMap.put("userid", strUID);
        localHashMap.put("password", strPW);        

        aq.ajax(strHostName, localHashMap, JSONObject.class, new AjaxCallback<JSONObject>()
        {
            public void callback(String paramString, JSONObject paramJSONObject, AjaxStatus paramAjaxStatus)
            {
                    if(paramJSONObject != null)
                    {
                        Log.d("LoginSignup", "LoginProcess call try : " + paramAjaxStatus.getCode() + " | " + paramJSONObject.toString());
                        return;
                    }
                    else
                    {
                        Log.d("LoginSignup", "ERROR : " + paramString + "|" + paramAjaxStatus.getCode() + "|" + paramAjaxStatus.getMessage());
                        return;                     
                    }


            }
        });

服务器PHP代码

 $arrMine = $db->rawQuery($arrQuery);

    //echo json_encode($arrMine);

    // Only One
    for($i=0;$i<count($arrMine);$i++){
      $objResult->usn = $arrMine[$i]['usn'];
      $objResult->userid = $arrMine[$i]['userid'];
      $objResult->password = $arrMine[$i]['password'];
      break;
    }

    //print_r($objResult);

    echo json_encode($arrNotice);

  } catch(Exception $e){
    echo json_encode($e->getMessage());
  }

错误码

错误：http://www.aropasoft.com/adpocket/member/login|-103|transform错误

我不知道为什么会出错……

Answer 1

试一试

 aq.ajax(strHostName, localHashMap, String.class, new AjaxCallback<String>()

然后从字符串中解析json

Answer 2

查询需要JsonObject。更改此行

echo json_encode($arrNotice);

为此：

echo json_encode($arrNotice, JSON_FORCE_OBJECT);