使用cakephp api在android上发布数据后启用获取响应

Question

我使用cakephp开发了一个api，现在我正在尝试在android上实现POST和GET功能。 GET工作正常但是当我发布数据时，它被添加到数据库但json响应不起作用！这是我发布数据时得到的结果： error message

这是我的android代码：

public class TaskManager {

    public static class AddLocation extends AsyncTask<String, Void, String> {
        String result ;

        @Override
        protected String doInBackground(String... params) {

            String addWaypointUrl = "http://xxx.xxx.xxx.xxx:xxxx/locations/addlocation";
            String token = "eyJ0eXAiOiJKV1QiLCJhbGciOiJIUzI1NiJ9.eyJzdWIiOjIsImV4cCI6MTUwMjI0MDc5NX0.3gRtjvVuMPJKJwvkzPcjK19zqJadCUiT2LH-Ut4TVgk";
            URL object = null;
            try {
                object = new URL(addWaypointUrl);
                HttpURLConnection con = (HttpURLConnection) object.openConnection();
                con.setDoOutput(true);
                con.setDoInput(true);
                con.setRequestProperty("Content-Type", "application/json");
                con.setRequestProperty("Accept", "application/json");
                con.setRequestProperty("Authorization", token);
                con.setRequestMethod("POST");
                JSONObject cred = new JSONObject();
                cred.put("id_user","12");
                cred.put("name","doci vida");
                cred.put("street","rue de paris");
                cred.put("description","restaurant italien");

                OutputStreamWriter wr = new OutputStreamWriter(con.getOutputStream());
                wr.write(cred.toString());
                wr.flush();
                StringBuilder sb = new StringBuilder();
                int HttpResult = con.getResponseCode();
                if (HttpResult == HttpURLConnection.HTTP_OK) {
                    BufferedReader br = new BufferedReader(
                            new InputStreamReader(con.getInputStream(), "utf-8"));
                    String line = null;
                    while ((line = br.readLine()) != null) {
                        sb.append(line + "\n");
                    }
                    br.close();
//                    JSONObject jObj = new JSONObject(sb.toString());
                    Log.e("result","f"+sb.toString());
                    if (sb.toString() != null) {
                        return sb.toString();
                    }
                } else {
                    System.out.println(con.getResponseMessage());
                }
            } catch (MalformedURLException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }catch (JSONException e) {
                e.printStackTrace();
            }
            return null;
        }


        @Override
        protected void onPostExecute(String result) {
            Log.e("result","null "+result);
        }
    }
}

我在发布问题之前在互联网上搜索了很多，但我无法解决这个问题。任何想法？

更新 这是我的PHP代码：

public function addLocation()
{
    $location = $this->Locations->newEntity();
    if ($this->request->is('post')) {
        $location = $this->Locations->patchEntity($location, $this->request->data);
        if ($this->Locations->save($location)) {
            $this->set([
                           'success' => 1,
                           '_serialize' => ['success']
                           ]);
        } else {
            $this->set([
                           'success' => 0,
                           'errors' => $location->errors(),
                           '_serialize' => ['success','errors']
                           ]);
        }
    }
}