Home | Contact Us | FAQ | Search & Site Map | Link to Us
Sign In | Join | Other 45 Sites in Network
HomeAnnouncementsWhite Papers
Discussion GroupsFirst AidDatabasesJavaBeansGUIJava 3DVirtual MachineCORBASecurityToolsGeneral
Java DirectoryOpen Source ProjectsSample Book ChaptersUser GroupsWeb Resources
Related Topics
Databases.NETMore Topics ...

Java Forum / General / October 2006

Tip: Looking for answers? Try searching our database.

URLConnection

Thread view: 
rplank@gmail.com - 18 Oct 2006 19:12 GMT
I am new to URL Connections so please forgive me if this is a stupid
question.

I am writing an Frame app that I want to use as a Client interface to a
web site that is writen in php, I am trying to figure out how to pass
the user name and password to this web site. if someone could point me
in the right driection that would be great.

Rob
Manish Pandit - 18 Oct 2006 19:33 GMT
Hi,

You need to find out the login form fields for the site, as well as the
URL to post this form to. You also need to check if the site requires
cookie handling at the client. To start with, you can use the example
provided here:

http://martin.nobilitas.com/java/cookies.html

It has an example of a form post and cookie-handling.

-cheers,
Manish
rplank@gmail.com - 19 Oct 2006 06:50 GMT
This was helpfull, I have an idea how to do it now but I can't seem to
figure out how to specify the web page with out getting "unreported
exception java.net.MalformedURLException; must be caught or declared to
be thrown" error

if i remove the URL line the program complies with out error

error -> URL url = new URL("http://www.kingsofchaos.com");
        String user="", email="", pass="",body1="";
        user = UserText.getText();
        email = EmailText.getText();
        pass = PasswordText.getText();

        body1 = ("field1="+user+"&field2="+email+"&field3="+pass+"");
        outputTA.setText(body1);
~~~~~~~going to call function~~~~~

static public String getURLPostString(URL url, String body) {
       StringBuffer sb = new StringBuffer();

    // find the newline character(s) on the current system
       String newline = null;
       try {
           newline = System.getProperty("line.separator");
       } catch (Exception e) {
           newline = "\n";
       }

       try {
       // URL must use the http protocol!
           HttpURLConnection conn = (HttpURLConnection)
url.openConnection();
           conn.setRequestMethod("POST");
           conn.setAllowUserInteraction(false); // you may not ask the
user
           conn.setDoOutput(true); // we want to send things
           // the Content-type should be default, but we set it anyway
           conn.setRequestProperty( "Content-type",
"application/x-www-form-urlencoded" );
           // the content-length should not be necessary, but we're
cautious
           conn.setRequestProperty( "Content-length",
Integer.toString(body.length()));

           // get the output stream to POST our form data
           OutputStream rawOutStream = conn.getOutputStream();
           PrintWriter pw = new PrintWriter(rawOutStream);

           pw.print(body); // here we "send" our body!
           pw.flush();
           pw.close();

           // get the input stream for reading the reply
           // IMPORTANT! Your body will not get transmitted if you get
the
           // InputStream before completely writing out your output
first!
           InputStream  rawInStream = conn.getInputStream();

           // get response
           BufferedReader rdr = new BufferedReader(new
InputStreamReader(rawInStream));
           String line;

           while ((line = rdr.readLine()) != null) {
               sb.append(line);
               sb.append(newline);
           }
           return sb.toString();
       } catch (Exception e) {
       System.out.println("Exception "+e.toString());
       e.printStackTrace();
    }
       return ""; // an exception occurred
   }

Rob

> Hi,
>
[quoted text clipped - 9 lines]
> -cheers,
> Manish
Andrew Thompson - 19 Oct 2006 07:39 GMT
> This was helpfull,

Please refrain from top-posting.  It makes threads very confusing.

>..I have an idea how to do it now but I can't seem to
> figure out how to specify the web page with out getting "unreported
> exception java.net.MalformedURLException; must be caught or declared to
> be thrown" error

comp.lang.java.help is a good group for beginners.

> if i remove the URL line the program complies with out error

Or, if you accept either of the options specified.. (cont. *)

2) 'declared'

 public URL getURL(String path) throws MalformedURLException {
    // the code that calls for a new URL
    return new URL(path);
 }

..but as soon as you go to call that method, you will end
with the same error from the code that *calls* it, which leads
to option ..1.
1) 'caught'

 public URL getURL(String path) {
   URL url = null;
   try {
      // the code that calls for a new URL
      url = new URL(path);
   } catch(MalformedURLException murle) { //exception caught!
      murle.printStackTrace();
   }
   return url;
 }

* ..the program will compile just fine (so long as
nothing else is wrong with the code).

Andrew T.


Free Magazines

Get these publications absolutely FREE for up to 12 months. There are no hidden fees and no obligation. Simply choose a title, complete the application form and submit it. Read more ...

Oracle MagazineNetwork ComputingComputer WorldBio-IT WorldeWeekInformation WeekInfosecurity
 
Sign In
Join
My Latest Posts
My Monitored Threads
My Blog
My Photo Gallery
My Profile
My Homepage

Start New Thread
Enable EMail Alerts
Rate this Thread



©2009 Advenet LLC   Privacy Policy - Terms of Use
This website includes both content owned or controlled by Advenet as well as content owned or controlled by third parties.