You're not echoing anything back to the browser.

For example change this:

if ($action == "logout")
  unset ($login_user);
  unset ($login_password);
  unset ($action);

to this:

if ($action == "logout")
  echo "You chose to log out<br>";
  unset ($login_user);
  unset ($login_password);
  unset ($action);
  echo "Thanks, you are now logged out.";

It might also help prevent confusion if you indented your code for each level of
curly brackets "{ }" you use.  Having all your brackets flush left is tough to

kind regards,

bill hollett

Kenneth Groth wrote:

> Greetings, all :) I'm looking for a PHP god, or at least someone who has the
> answer to my problem.  I have a script for a login system that can handle
> multiple users at (
> is the source code, some may
> have to click 'view source' in their browsers) - but it does not work.try
> it-use handle "new_handle" with password "somepwd" ...that's one of the
> things in the MySQL database.  oh, the login works great.  but those options
> in the select box don't - both take me to a nasty white screen of death when
> I try to log out or change password ....the coding appears OK to a friend of
> mine and excellent PHP coder.  Is there something I missed?
> -Stoney
> P.S. - I know that the mysql_connect () is not given a password...because,
> after all, I'm not putting in my MySQL database pass in there :P

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