JSONArray对Volley for Android的回应

I am posting some data into the database using Volley and I get the following jsonarray response.

我使用Volley将一些数据发布到数据库中，我得到以下jsonarray响应。

[
  {
  "nickname":"panikos",
  "username":"panikos@gmail.com",
  "user_type":"LEADER",
  "latest_steps":"0"
  }
]

This is a sample of my code that unfortunately doesn't log out or debug the variable of "nickname" object:(.

这是我的代码示例，遗憾的是它没有注销或调试“nickname”对象的变量:(。

final JsonArrayRequest jsonObjReq1 = new 
JsonArrayRequest(AppConfig.URL_GET_TEAM, jsonObject,
            new com.android.volley.Response.Listener<JSONArray>() {

                @Override
                public void onResponse(JSONArray response) {
                    Log.d("TAG", response.toString());

                    try {
                        JSONArray jsonArray = new JSONArray(response);

                        for(int i=0;i<jsonArray.length();i++){
                            JSONObject jresponse = 
                jsonArray.getJSONObject(i);
                String nickname =                                 
           jresponse.getString("nickname");
                            Log.d("nickname",nickname);
                        }
                    } catch (JSONException e) {
                        e.printStackTrace();
                    }
                    //pDialog.dismiss();

                }
            }, new com.android.volley.Response.ErrorListener() {

        @Override
        public void onErrorResponse(VolleyError error) {
            VolleyLog.d("TAG", "Error: " + error.getMessage());
            //pDialog.dismiss();

        }
    }) {

        @Override
        public String getBodyContentType() {
            return "application/json; charset=utf-8";
        }


    };

Any ideas? Am I missing something?

有任何想法吗？我错过了什么吗？

Thanks.

谢谢。

2 个解决方案

#1

I the problem might be - you are already getting response as a JSONArray.

我的问题可能是 - 你已经得到了JSONArray的响应。

So, you can Call

所以，你可以打电话

JSONObject jresponse = response.getJSONObject(0);

JSONObject jresponse = response.getJSONObject（0）;

and if you have more than 1 object in response, then

如果您有多个对象作为响应，那么

for(int i = 0; i < response.length(); i++){
    JSONObject jresponse = response.getJSONObject(i);
    String nickname = jresponse.getString("nickname");
    Log.d("nickname", nickname);
}

Remove this :

删除这个：

               try {
                    JSONArray jsonArray = new JSONArray(response);

                    for(int i=0;i<jsonArray.length();i++){
                        JSONObject jresponse = 
            jsonArray.getJSONObject(i);
            String nickname =                                 
       jresponse.getString("nickname");
                        Log.d("nickname",nickname);
                    }
                } catch (JSONException e) {
                    e.printStackTrace();
                }

and add :

并添加：

try {
    JSONObject jresponse = response.getJSONObject(0);
    String nickname = jresponse.getString("nickname");
    Log.d("nickname",nickname);
}catch (JSONException e) {
    e.printStackTrace();
}

The Code looks good, however i think you might be missing a call to add jsonObjReq1 in the request queue. I would suggest to use Singleton Pattern.

代码看起来不错，但我认为你可能会错过在请求队列中添加jsonObjReq1的调用。我建议使用Singleton Pattern。

#2

Fixed!!!

固定！！！

                @Override
                public void onResponse(JSONArray response) {
                    Log.d("TAG", response.toString());

                    try {

                        Log.d("JsonArray",response.toString());
                        for(int i=0;i<response.length();i++){
                            JSONObject jresponse = response.getJSONObject(i);
                            String nickname = jresponse.getString("nickname");
                            Log.d("nickname",nickname);
                        }
                    } catch (JSONException e) {
                        e.printStackTrace();
                    }
                    //pDialog.dismiss();

                }

There was no need to create a new JSONArray. It was created inside the onResponse() method. The next project I am assigned to do is going to have more complicate webservices.omg!!!

无需创建新的JSONArray。它是在onResponse（）方法中创建的。我被指派做的下一个项目将会有更复杂的webservices.omg！

#1