Java - sending HTTP parameters via POST method easily

前端 未结 17 1737
借酒劲吻你
借酒劲吻你 2020-11-21 05:54

I am successfully using this code to send HTTP requests with some parameters via GET method

void sendRequest(String request)
{
            


        
17条回答
  •  一生所求
    2020-11-21 06:43

    here i sent jsonobject as parameter //jsonobject={"name":"lucifer","pass":"abc"}//serverUrl = "http://192.168.100.12/testing" //host=192.168.100.12

      public static String getJson(String serverUrl,String host,String jsonobject){
    
        StringBuilder sb = new StringBuilder();
    
        String http = serverUrl;
    
        HttpURLConnection urlConnection = null;
        try {
            URL url = new URL(http);
            urlConnection = (HttpURLConnection) url.openConnection();
            urlConnection.setDoOutput(true);
            urlConnection.setRequestMethod("POST");
            urlConnection.setUseCaches(false);
            urlConnection.setConnectTimeout(50000);
            urlConnection.setReadTimeout(50000);
            urlConnection.setRequestProperty("Content-Type", "application/json");
            urlConnection.setRequestProperty("Host", host);
            urlConnection.connect();
            //You Can also Create JSONObject here 
            OutputStreamWriter out = new OutputStreamWriter(urlConnection.getOutputStream());
            out.write(jsonobject);// here i sent the parameter
            out.close();
            int HttpResult = urlConnection.getResponseCode();
            if (HttpResult == HttpURLConnection.HTTP_OK) {
                BufferedReader br = new BufferedReader(new InputStreamReader(
                        urlConnection.getInputStream(), "utf-8"));
                String line = null;
                while ((line = br.readLine()) != null) {
                    sb.append(line + "\n");
                }
                br.close();
                Log.e("new Test", "" + sb.toString());
                return sb.toString();
            } else {
                Log.e(" ", "" + urlConnection.getResponseMessage());
            }
        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        } catch (JSONException e) {
            e.printStackTrace();
        } finally {
            if (urlConnection != null)
                urlConnection.disconnect();
        }
        return null;
    }
    

提交回复
热议问题