Olivier Mueller a écrit :
got this on a private mail account: (no time to look
at it now,
will check during the week).
regards,
Olivier
-------- Forwarded Message --------
From: Ing. Jozef Sakalos <jsakalos(a)aariadne.com>
Subject: Better logout
Date: Sun, 18 Apr 2004 19:37:05 +0200
Mailer: KMail/1.6.51
Hi Olivier,
I just downloaded and installed CVS version of phpMyAdmin and I found
that you've added the logout button to the left frame.
The drawback is that if you click it you cannot login with the same name
and password. You've probably used example from PHP documentation.
He is talking about 'http' auth_type only. This is the old limitation
documented in FAQ 3.1.
I've been working on the logout feature too, and I
found the better
solution (see code snippet at the end of this e-mail).
I use logout button that POSTs cmd=logout to the PHP code. When logout
command is processed I set cookie (well, this is browser settings
dependent) to prevent infinite loop of log-in dialogs.
If the browser needs to accept cookies for this, I am not sure
it's a good general solution.
Anyway, he should not just check $logging_out but use $_COOKIE
and friends.
Marc
Once the user authenticates, I delete the cookie.
This way I can log-in with the same name/password.
Use this if you want and let me know your opinion.
Best Regards,
Ing. Jozef Sakalos
jsakalos(a)aariadne.com
<?
if(!isset($_SERVER["PHP_AUTH_USER"]) || ("logout" == $cmd &&
!
$logging_out)) {
if("logout" == $cmd) {
setcookie("logging_out", "1");
}
header('WWW-Authenticate: Basic realm="Theseus"');
header('HTTP/1.0 401 Unauthorized');
header("Status: 401 Unauthorized");
echo "You have to authenticate<br>";
exit;
}
setcookie("logging_out", "", time() - 3600);
echo "Hello: $PHP_AUTH_USER<br>";
echo "Your password is: $PHP_AUTH_PW";
echo <<<eot
<html><head></head><body>
<br>
<br>
<form action="$PHP_SELF" method="post">
<input type="hidden" name="cmd" value="logout">
<input type="submit" value="Logout">
</form>
cmd=$cmd
eot;
echo "</body></html>";
?>
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