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.