Как передать объект JSON в запрос POST HttpURLConnection

Ответ 1

Попробуйте

...
httpCon.setRequestMethod("POST");
httpCon.connect(); // Note the connect() here
...
OutputStream os = httpCon.getOutputStream();
OutputStreamWriter osw = new OutputStreamWriter(os, "UTF-8");
...    
osw.write(msg.toString());
osw.flush();
osw.close();

для отправки данных.

для получения данных попробуйте:

BufferedReader br = new BufferedReader(new InputStreamReader( httpCon.getInputStream(),"utf-8"));
String line = null;
while ((line = br.readLine()) != null) {
    sb.append(line + "\n");
}
br.close();
System.out.println(""+sb.toString());

Ответ 2

HttpURLConnection является громоздким для использования. С DavidWebb, крошечной оберткой вокруг HttpURLConnection, вы можете написать ее вот так:

JSONObject msg;  //passed in as a parameter to this method

Webb webb = Webb.create();
JSONObject result = webb.post("http://my-url/path/to/res")
    .useCaches(false)
    .body(msg)
    .ensureSuccess()
    .asJsonObject()
    .getBody();

Если вам это не нравится, в приведенной ссылке есть список альтернативных библиотек.

Почему мы все должны писать один и тот же шаблонный код каждый день? Кстати, приведенный выше код более читабельен и менее подвержен ошибкам. HttpURLConnection имеет ужасный интерфейс. Это должно быть завернуто!

Ответ 3

 public String sendHTTPData(String urlpath, JSONObject json) {
        HttpURLConnection connection = null;
        try {
            URL url=new URL(urlpath);
            connection = (HttpURLConnection) url.openConnection();
            connection.setDoOutput(true);
            connection.setDoInput(true);
            connection.setRequestMethod("POST");
            connection.setRequestProperty("Content-Type", "application/json");
            connection.setRequestProperty("Accept", "application/json");
            OutputStreamWriter streamWriter = new OutputStreamWriter(connection.getOutputStream());
            streamWriter.write(json.toString());
            streamWriter.flush();
            StringBuilder stringBuilder = new StringBuilder();
            if (connection.getResponseCode() == HttpURLConnection.HTTP_OK){
                InputStreamReader streamReader = new InputStreamReader(connection.getInputStream());
                BufferedReader bufferedReader = new BufferedReader(streamReader);
                String response = null;
                while ((response = bufferedReader.readLine()) != null) {
                    stringBuilder.append(response + "\n");
                }
                bufferedReader.close();

                Log.d("test", stringBuilder.toString());
                return stringBuilder.toString();
            } else {
                Log.e("test", connection.getResponseMessage());
                return null;
            }
        } catch (Exception exception){
            Log.e("test", exception.toString());
            return null;
        } finally {
            if (connection != null){
                connection.disconnect();
            }
        }
    }`

назовите этот метопд в doitbackground в asynctask

Ответ 4

Следуйте следующему примеру:

public static PricesResponse getResponse(EventRequestRaw request) {

    // String urlParameters  = "param1=a&param2=b&param3=c";
    String urlParameters = Piping.serialize(request);

    HttpURLConnection conn = RestClient.getPOSTConnection(endPoint, urlParameters);

    PricesResponse response = null;

    try {
        // POST
        OutputStreamWriter writer = new OutputStreamWriter(conn.getOutputStream());
        writer.write(urlParameters);
        writer.flush();

        // RESPONSE
        BufferedReader reader = new BufferedReader(new InputStreamReader((conn.getInputStream()), StandardCharsets.UTF_8));
        String json = Buffering.getString(reader);
        response = (PricesResponse) Piping.deserialize(json, PricesResponse.class);

        writer.close();
        reader.close();

    } catch (Exception e) {
        e.printStackTrace();
    }

    conn.disconnect();

    System.out.println("PricesClient: " + response.toString());

    return response;
}


public static HttpURLConnection getPOSTConnection(String endPoint, String urlParameters) {

    return RestClient.getConnection(endPoint, "POST", urlParameters);

}


public static HttpURLConnection getConnection(String endPoint, String method, String urlParameters) {

    System.out.println("ENDPOINT " + endPoint + " METHOD " + method);
    HttpURLConnection conn = null;

    try {
        URL url = new URL(endPoint);
        conn = (HttpURLConnection) url.openConnection();
        conn.setRequestMethod(method);
        conn.setDoOutput(true);
        conn.setRequestProperty("Content-Type", "text/plain");

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

    return conn;
}

Ответ 5

this без json Строковые сообщения на сервер

 class PostLogin extends AsyncTask<Void, Void, String> {
        @Override
        protected String doInBackground(Void... params) {
            String response = null;

            Uri.Builder builder= new Uri.Builder().appendQueryParameter("username","amit").appendQueryParameter("password", "amit");
            String parm=builder.build().getEncodedQuery();
      try
           {

               response = postData("your url here/",parm);
           }catch (Exception e)
           {
               e.printStackTrace();
           }
            Log.d("test", "response string is:" + response);
            return response;
        }
    }


private String postData(String path, String param)throws IOException {
        StringBuffer response = null;

        URL  url = new URL(path);
        HttpURLConnection  connection = (HttpURLConnection) url.openConnection();
        connection.setRequestMethod("POST");
        connection.setDoOutput(true);
//        connection.setRequestProperty("Content-Type", "application/json");
//        connection.setRequestProperty("Accept", "application/json");
            OutputStream out = connection.getOutputStream();
            out.write(param.getBytes());
            out.flush();
            out.close();
            int responseCode = connection.getResponseCode();
            if (responseCode == HttpURLConnection.HTTP_OK) {
                BufferedReader br = new BufferedReader(new InputStreamReader(connection.getInputStream()));
                String line;
                response = new StringBuffer();
                while ((line = br.readLine()) != null) {
                    response.append(line);
                }
                br.close();
            }

        return response.toString();
    }