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how to do an online xml request in java to a url that require authentication

I want to do an online xml request in java but the server responds with 401 error that means that there is an authentication that is need to access the server. I have the certfile.cer that i can use to do the authentication but i dont know how to load it in java.How can I achieve this in java? Here is part of my code.
StringBuilder answer = new StringBuilder();
URL url = new URL("www.myurl.com");
URLConnection conn = url.openConnection();
OutputStreamWriter writer = new OutputStreamWriter(conn.getOutputStream());
writer.write(xml);
writer.flush();
String line;
while ((line = reader.readLine()) != null)
{
answer.append(line);
}

How do i get Instagram Code for Oauth authorisation?

I am making desktop app in java for instagram right now and i don't know how to get code from callback url. My app is desktop so my REDIRECT-URI is http://instagram.com. So in my app i send a request to https://api.instagram.com/oauth/authorize/?client_id=CLIENT-ID&redirect_uri=REDIRECT-URI&response_type=code and get redirected to instagram.com?code=CODE. And also forgot to mention, I get a code once by my hands a i gave all permissions. So now then i make request i redirected straightly to URL with code.
How do i get code from callback uri in program in java? This is my code:
static void GetCode()
{
try {
String url = "https://api.instagram.com/oauth/authorize/?client_id=" + ClientId + "&redirect_uri=" + BackUri + "&response_type=code&scope=likes+comments+relationships";
HttpClient client = new DefaultHttpClient();
HttpGet request = new HttpGet(url);
HttpResponse response = client.execute(request);
BufferedReader rd = new BufferedReader(
new InputStreamReader(response.getEntity().getContent()));
StringBuffer result = new StringBuffer();
String line = "";
while ((line = rd.readLine()) != null) {
result.append(line);
}
System.out.println(result.toString());
}
catch(Exception e)
{
System.out.print("GET CODE ERROR->"+e.toString());
}
}

Change REDIRECT_URI to http://localhost. So the actual 'redirection' code should be done on your side. And you can parse value of argument code. Also look at the answers to this question. This will help you to grab URI with the code.

HTML Form Action Java

I have a java program which I want to input something into an html form. If possible it could just load a url like this
.../html_form_action.asp?kill=Kill+Server
But i'm not sure how to load a url in Java. How would I do this? Or is there a better way to send an action to an html form?

Depending on your security, you can make an HTTP call in Java. It is often referred to as a RESTFul call. The HttpURLConnection class offers encapsulation for basic GET/POST requests. There is also an HttpClient from Apache.

Here's how you can use URLConnection to send a simple HTTP request.
URL url = new URL(url + "?" + query);
// set connection properties
URLConnection connection = url.openConnection();
connection.setRequestProperty("Accept-Charset", "UTF-8");
connection.connect(); // send request
// read response
BufferedReader reader = new BufferedReader(
new InputStreamReader(connection.getInputStream()));
String line = null;
while ((line = reader.readLine()) != null) {
System.out.println(line);
}
reader.close(); // close connection

Getting URL after a redirect using HttpClient.Execute(HttpGet)

I have searched for a while and I am not finding a clear answer. I am trying to log into a webstie.
https://hrlink.healthnet.com/
This website redirects to a login page that is not consitent. I have to post my login credentials to the redirected URL.
Im am trying to code this in Java but I do not understand how to get the URL from the response. It may look a bit messy but I have it this way while I am testing.
HttpGet httpget = new HttpGet("https://hrlink.healthnet.com/");
HttpResponse response = httpclient.execute(httpget);HttpEntity entity = response.getEntity();
String redirectURL = "";
for(org.apache.http.Header header : response.getHeaders("Location")) {
redirectURL += "Location: " + header.getValue()) + "\r\n";
}
InputStream is;
is = entity.getContent();
BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
String result = sb.toString();
I know i get redirected because my result string shows be the actual login page but I am not able to get the new URL.
In FireFox I am using TamperData. When I navigate to this website https://hrlink.healthnet.com/ I have a GET with a 302 - Found and the Location of the Login Page. Then another GET to the actual Login Page
Any help is greatly appreciated thank you.

Check out w3c documentation:
10.3.3 302 Found
The temporary URI SHOULD be given by the Location field in the response. Unless the request method was HEAD, the entity of the response SHOULD contain a short hypertext note with a hyperlink to the new URI(s).
If the 302 status code is received in response to a request other than GET or HEAD, the user agent MUST NOT automatically redirect the request unless it can be confirmed by the user, since this might change the conditions under which the request was issued.
One solution is to use POST method to break auto-redirecting at client side:
HttpPost request1 = new HttpPost("https://hrlink.healthnet.com/");
HttpResponse response1 = httpclient.execute(request1);
// expect a 302 response.
if (response1.getStatusLine().getStatusCode() == 302) {
String redirectURL = response1.getFirstHeader("Location").getValue();
// no auto-redirecting at client side, need manual send the request.
HttpGet request2 = new HttpGet(redirectURL);
HttpResponse response2 = httpclient.execute(request2);
... ...
}
Hope this helps.

Imgur API request using Java returns 400 status

I am trying to send a GET request to the Imgur API to upload an image.
When I use the following code I receive a 400 status response from the Imgur server - which, according to the Imgur error documentation, means I am missing or have incorrect parameters.
I know the parameters are correct as I have tested them directly in the browser URL (which successfully uploads an image) - so I must not be adding the parameters correctly within the code:
private void addImage(){
String data = URLEncoder.encode("image", "UTF-8") + "=" + URLEncoder.encode("http://www.lefthandedtoons.com/toons/justin_pooling.gif", "UTF-8");
data += "&" + URLEncoder.encode("key", "UTF-8") + "=" + URLEncoder.encode("myPublicConsumerKey", "UTF-8");
// Send data
java.net.URL url = new java.net.URL("http://api.imgur.com/2/upload.json");
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) {
Logger.info( line );
}
wr.close();
rd.close();
}
This code is based on the API examples provided by Imgur.
Can anyone tell me what I am doing wrong and how I may resolve the problem?
Thanks.

In this sample, imgur service returns 400 Bad Request status response with a non-empty body because of incorrect API key. In case of non successful HTTP response you shold read the response body from an error input stream. For example:
// Get the response
InputStream is;
if (((HttpURLConnection) conn).getResponseCode() == 400)
is = ((HttpURLConnection) conn).getErrorStream();
else
is = conn.getInputStream();
BufferedReader rd = new BufferedReader(new InputStreamReader(is));
And, by the way your example is POST, not GET, because you are sending the parameters in the request body instead of the URL.