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HTTPS Post Request through the URL

Amitabh Sinha
Greenhorn

Joined: Sep 02, 2004
Posts: 1
Hi Guys

When I try sending the https post request through the java program

String param = "https://abcdefg.com";
URL url = new URL(param);
HttpsURLConnection httpConnection =
(HttpsURLConnection)url.openConnection();
jrunCookie = httpConnection.getHeaderField("set-cookie");
httpConnection.setUseCaches(false);
httpConnection.setRequestProperty("Content-Type","application/x-www-form-urlencoded");
httpConnection.setRequestProperty("referer", param);
HttpsURLConnection.setFollowRedirects(false);
httpConnection.setInstanceFollowRedirects(false);
httpConnection.setDoInput(true);
httpConnection.setDoOutput(true);
httpConnection.setRequestMethod("POST");



it throws the exception and exception reads as URL is not correct. it should be abcdef.com

Could sanyone help me with the code forthis perpose.
Pat Hays
Ranch Hand

Joined: Aug 20, 2004
Posts: 138
have a look the following code, it may be helpful for you.
Sending a POST Request Using a Socket.

try {
// Construct data
String data = URLEncoder.encode("key1", "UTF-8") + "=" + URLEncoder.encode("value1", "UTF-8");
data += "&" + URLEncoder.encode("key2", "UTF-8") + "=" + URLEncoder.encode("value2", "UTF-8");

// Send data
URL url = new URL("http://hostname:80/cgi");
URLConnection conn = url.openConnection();
conn.setDoOutput(true);
OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
wr.write(data);
wr.flush();

// Get the response
BufferedReader rd = new BufferedReader(new InputStreamReader(conn.getInputStream()));
String line;
while ((line = rd.readLine()) != null) {
// Process line...
}
wr.close();
rd.close();
} catch (Exception e) {
}


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