1. Just pass uzeriz to the function, not session("uzeriz");
2. FYI: you don't need this construct: while($row =
mysql_fetch_row($result)){
Based upon the previous tests, there is ONLY 1 entry. Therefore, go with:
$row = mysql_fetch_row($result);
-----Original Message-----
From: Mantas Kriauciunas [mailto:[EMAIL PROTECTED]]
Sent: Monday, March 25, 2002 6:39 AM
To: PHP General List
Subject: [PHP] [newbie] Something is messed up, anybody can help?
Hey PHP General List,
Intro:
I am creating small login script and few things doesn't work right.
Code is hare:
admin.php
if(isset($subm_login))
{
$session["logged"]=0;
$session["uzeriz"]="";
$result = mysql_query("select * from uzer where user='$uzr_name' and
pass='$uzr_passwd'");
if(!$result) {
echo "This error should not be hare. But it sayes that it
could not search for your user";
} else {
if(mysql_num_rows($result)!=1) {
echo "Login Not Found";
} else {
while($row = mysql_fetch_row($result)){
$session["uzeriz"]=$row["user"];
$session["logged"]=1;
$session["pass"]=$row["pass"];
}
}
}
session_register("session");
}
if( $session["logged"]==1 )
{
draw_admin_menu($session["uzeriz"]);
}
if( $session["logged"]==0 )
{
draw_login_box();
}
--------------
adm_func.php (this file is included in admin.php)
function draw_admin_menu($uzer)
{
global $session;
echo "<p>User Logged In</p>\n";
echo "<p>--------------</p>\n";
echo "<p>$uzer</P>\n";
}
------------------------------
Problem:
the problem is that i cant see $uzer its empty. nothing is
there....what i am doing wrong? all table names and everything is
correct but i get empty thing.
If that is dumb question im sorry :) i am kinda new in this :)
:------------------------------:
Have A Nice Day!
Mantas Kriauciunas A.k.A mNTKz
Contacts:
[EMAIL PROTECTED]
Http://mntkz-hata.visiems.lt
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
