I'm using the code below to send an http POST request which sends an object to a WCF service. This works ok, but what happens if my WCF service needs also other parameters? How can I send them from my Android client?
This is the code I've written so far:
StringBuilder sb = new StringBuilder();
String http = "http://android.schoolportal.gr/Service.svc/SaveValues";
HttpURLConnection urlConnection=null;
try {
URL url = new URL(http);
urlConnection = (HttpURLConnection) url.openConnection();
urlConnection.setDoOutput(true);
urlConnection.setRequestMethod("POST");
urlConnection.setUseCaches(false);
urlConnection.setConnectTimeout(10000);
urlConnection.setReadTimeout(10000);
urlConnection.setRequestProperty("Content-Type","application/json");
urlConnection.setRequestProperty("Host", "android.schoolportal.gr");
urlConnection.connect();
//Create JSONObject here
JSONObject jsonParam = new JSONObject();
jsonParam.put("ID", "25");
jsonParam.put("description", "Real");
jsonParam.put("enable", "true");
OutputStreamWriter out = new OutputStreamWriter(urlConnection.getOutputStream());
out.write(jsonParam.toString());
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();
System.out.println(""+sb.toString());
}else{
System.out.println(urlConnection.getResponseMessage());
}
} catch (MalformedURLException e) {
e.printStackTrace();
}
catch (IOException e) {
e.printStackTrace();
} catch (JSONException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}finally{
if(urlConnection!=null)
urlConnection.disconnect();
}
Harish
Posting parameters Using POST:-
URL url;
URLConnection urlConn;
DataOutputStream printout;
DataInputStream input;
url = new URL (getCodeBase().toString() + "env.tcgi");
urlConn = url.openConnection();
urlConn.setDoInput (true);
urlConn.setDoOutput (true);
urlConn.setUseCaches (false);
urlConn.setRequestProperty("Content-Type","application/json");
urlConn.setRequestProperty("Host", "android.schoolportal.gr");
urlConn.connect();
//Create JSONObject here
JSONObject jsonParam = new JSONObject();
jsonParam.put("ID", "25");
jsonParam.put("description", "Real");
jsonParam.put("enable", "true");
The part which you missed is in the the following... i.e., as follows..
// Send POST output.
printout = new DataOutputStream(urlConn.getOutputStream ());
printout.writeBytes(URLEncoder.encode(jsonParam.toString(),"UTF-8"));
printout.flush ();
printout.close ();
The rest of the thing you can do it.
Amir Qayyum Khan
try some thing like blow:
SString otherParametersUrServiceNeed = "Company=acompany&Lng=test&MainPeriod=test&UserID=123&CourseDate=8:10:10";
String request = "http://android.schoolportal.gr/Service.svc/SaveValues";
URL url = new URL(request);
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setDoOutput(true);
connection.setDoInput(true);
connection.setInstanceFollowRedirects(false);
connection.setRequestMethod("POST");
connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
connection.setRequestProperty("charset", "utf-8");
connection.setRequestProperty("Content-Length", "" + Integer.toString(otherParametersUrServiceNeed.getBytes().length));
connection.setUseCaches (false);
DataOutputStream wr = new DataOutputStream(connection.getOutputStream ());
wr.writeBytes(otherParametersUrServiceNeed);
JSONObject jsonParam = new JSONObject();
jsonParam.put("ID", "25");
jsonParam.put("description", "Real");
jsonParam.put("enable", "true");
wr.writeBytes(jsonParam.toString());
wr.flush();
wr.close();
References :
来源：https://stackoverflow.com/questions/13911993/sending-a-json-http-post-request-from-android