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
read.
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 http://www.stoneneedle.f2s.com/loginsystem2.php3 (
> http://www.stoneneedle.f2s.com/loginsystem2.txt 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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