его НЕ ДУПЛИКАТ. Ссылка, которая была предоставлена, является СТАРОЙ. "http client" был удален в api23

Я хочу отправить объект json:

{"emailId":"ashish.bhatt@mobimedia.in","address":"Naya bans","city":"Noida","pincode":"201301","account_number":"91123546374208","bank_name":"Axis Bank","branch_name":"91123546374208","ifsc_code":"UTI0000879"}

На URL:

http://10digimr.mobimedia.in/api/mobile_retailer/update_profile Как это сделать? через почтовый метод?

МЕТОД:

 POST /api/mobile_retailer/update_profile

ОБЯЗАТЕЛЬНЫЙ КЛЮЧ:

{"emailId","address"}

ЗАПРОСИТЬ JSON:

{"emailId":"ashish.bhatt@mobimedia.in","address":"Naya bans","city":"Noida","pincode":"201301","account_number":"91123546374208","bank_name":"Axis Bank","branch_name":"91123546374208","ifsc_code":"UTI0000879"}

ОТВЕТ:

{"message":"Mail Send","data":true,"status":200}
1
user6092109 25 Апр 2016 в 09:54

2 ответа

Лучший ответ

Определите класс AsyncT и вызовите его в методе onCreate, используя:

AsyncT asyncT = new AsyncT();
asyncT.execute();

Определение класса:

class AsyncT extends AsyncTask<Void,Void,Void>{

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

            try {
                URL url = new URL(""); //Enter URL here
                HttpURLConnection httpURLConnection = (HttpURLConnection)url.openConnection();
                httpURLConnection.setDoOutput(true);
                httpURLConnection.setRequestMethod("POST"); // here you are telling that it is a POST request, which can be changed into "PUT", "GET", "DELETE" etc.
                httpURLConnection.setRequestProperty("Content-Type", "application/json"); // here you are setting the `Content-Type` for the data you are sending which is `application/json`
                httpURLConnection.connect();

                JSONObject jsonObject = new JSONObject();
                jsonObject.put("para_1", "arg_1");

                DataOutputStream wr = new DataOutputStream(httpURLConnection.getOutputStream());
                wr.writeBytes(jsonObject.toString());
                wr.flush();
                wr.close();

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

            return null;
        }


    }
5
Michał Perłakowski 15 Май 2016 в 07:26

@Sandip Subedi Вот как вы получаете ответ от httpURLConnection

class AsyncT extends AsyncTask<Void,Void,Void>{

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

        try {
            URL url = new URL(""); //Enter URL here
            HttpURLConnection httpURLConnection = (HttpURLConnection)url.openConnection();
            httpURLConnection.setDoOutput(true);
            httpURLConnection.setRequestMethod("POST"); // here you are telling that it is a POST request, which can be changed into "PUT", "GET", "DELETE" etc.
            httpURLConnection.setRequestProperty("Content-Type", "application/json"); // here you are setting the `Content-Type` for the data you are sending which is `application/json`
            httpURLConnection.connect();

            JSONObject jsonObject = new JSONObject();
            jsonObject.put("para_1", "arg_1");

            DataOutputStream wr = new DataOutputStream(httpURLConnection.getOutputStream());
            wr.writeBytes(jsonObject.toString());
            wr.flush();
            wr.close();

            InputStream response = httpURLConnection.getInputStream();
            BufferedReader reader = new BufferedReader(newInputStreamReader(response);
            StringBuilder sb = new StringBuilder();
            String line = null;
            try {
                while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
                }
            } catch (IOException e) {
                e.printStackTrace();
            } finally {
                try {
                   is.close();
                } catch (IOException e) {
                   e.printStackTrace();
                }
            }

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

        return null;
    }


}
0
terminator 12 Апр 2019 в 12:21