View Full Version : [php]Little bit of help! Why doesnt this work!
Hey!
So this is the third time ive come for help in 2 days and i feel bad :L So im more than understanding if people dont want to help me out!
Ive got this code:
<?
session_start();
include ("config.php");
$username = $_SESSION['username'];
$logged = mysql_query("SELECT * FROM `users` WHERE `username`='$username'");
$logged = mysql_fetch_array($logged);
?>
<head>
<link href="style.css" rel="stylesheet" type="text/css" />
</head>
<?
if($logged["rank"] != "1") {
// not an admin.
exit("You need to be an administrator to view this feature"); // you can redirect to an error page, include something here, doesn't matter, just exit.
}
// aha.. they are an admin.. show the content.
include ("menus/anav.php")
?>
Dentafrice helped me with some of it but i went on to play around alot with it and see what i could get out of it!!
Now it doesnt work for some reason and nothing appears when logged in as rank 1 or any other ranks....
Can anyone help me!! THANKS :)
Edited by ReviewDude (Forum Moderator): Identical threads merged.
Joshh
06-04-2009, 03:54 PM
Hey!
So this is the third time ive come for help in 2 days and i feel bad :L So im more than understanding if people dont want to help me out!
Ive got this code:
<?
session_start();
include ("config.php");
$username = $_SESSION['username'];
$logged = mysql_query("SELECT * FROM `users` WHERE `username`='$username'");
$logged = mysql_fetch_array($logged);
?>
<head>
<link href="style.css" rel="stylesheet" type="text/css" />
</head>
<?
if($logged["rank"] != "1") {
// not an admin.
exit("You need to be an administrator to view this feature"); // you can redirect to an error page, include something here, doesn't matter, just exit.
}
// aha.. they are an admin.. show the content.
include ("menus/anav.php")
?>Dentafrice helped me with some of it but i went on to play around alot with it and see what i could get out of it!!
Now it doesnt work for some reason and nothing appears when logged in as rank 1 or any other ranks....
Can anyone help me!! THANKS :)
Edited by ReviewDude (Forum Moderator): Identical threads merged.
I'm pretty sure it's for this reason, although I could be making myself look stupid.
If they are an admin then it shows the content but the content you're showing is
"include ("menus/anav.php")" but that doesn't show it it just includes it
I suggest echoing it for example:
echo ("contents of menus/anav.php here");
tried, didnt work :S
Lol :L DW ill fix :) thanks for your help anyway!
Dentafrice
06-04-2009, 04:17 PM
Try this and tell me what comes up:
<?
session_start();
include ("config.php");
$username = $_SESSION['username'];
echo "Username: {$username}<br />";
$logged = mysql_query("SELECT * FROM `users` WHERE `username`='$username'");
$logged = mysql_fetch_array($logged);
echo "<pre>";
print_r($logged);
echo "</pre>";
?>
Joshh
06-04-2009, 04:38 PM
I think it's because you're defining if they're not admin to not let them in but you're not defining if they're logged in, I would of used something like this, but it's probbably not like you need but it's worth a shot:
<?
session_start();
include ("config.php");
$username = $_SESSION['username'];
$logged = mysql_query("SELECT * FROM `users` WHERE `username`='$username'");
$logged = mysql_fetch_array($logged);
?>
<head>
<link href="style.css" rel="stylesheet" type="text/css" />
</head>
<?
if($logged["rank"] != "1") {
// not an admin.
exit("You need to be an administrator to view this feature"); // you can redirect to an error page, include something here, doesn't matter, just exit.
}
else
{
if($logged["rank"] != "admin rank here") {
// is an admin.
exit("You're an admin!"); // you can redirect to an error page, include something here, doesn't matter, just exit.
}
echo ("the contents of menus/anav.php here")
}
?> then again probbably wrong again lol i'm just thinking through of what it could be. There's probbably an error in there, wasn't really paying attention when did it.
Dentafrice
06-04-2009, 04:49 PM
What is with the logic there? You're checking to see if they're not an admin, then you're checking to see if they are not an admin again..?
Joshh
06-04-2009, 08:13 PM
What is with the logic there? You're checking to see if they're not an admin, then you're checking to see if they are not an admin again..?
no because
if($logged["rank"] != "1") {
// not an admin.
exit("You need to be an administrator to view this feature"); // you can redirect to an error page, include something here, doesn't matter, just exit.
}
checks to see if they're not an admin, but it's not checking to see if they are an admin so I added one to check if they are admin, so if they are admin it displays the anav.php if theyre not it displays the "you're not an admin" message
Dentafrice
06-04-2009, 08:19 PM
no because
checks to see if they're not an admin, but it's not checking to see if they are an admin so I added one to check if they are admin, so if they are admin it displays the anav.php if theyre not it displays the "you're not an admin" message
*Removed*
This is your code..
This checks to see if the rank is not 1, which I believe is the administrator rank.
If rank is not 1 (admin rank) Then display the message. You don't need the else, it exits.. so you don't need to else it.
if($logged["rank"] != "1") {
// not an admin.
exit("You need to be an administrator to view this feature"); // you can redirect to an error page, include something here, doesn't matter, just exit.
}
Now why *Removed* are you doing the same thing?
If rank is not admin rank Then display "You are an admin!" *Removed* You're saying they are an admin when they are not. Plus.. if they are not an admin.. it's not even going to get to this section anyway.
if($logged["rank"] != "admin rank here") {
// is an admin.
exit("You're an admin!"); // you can redirect to an error page, include something here, doesn't matter, just exit.
}
echo ("the contents of menus/anav.php here")
*Removed*
Edited by ReviewDude (Forum Moderator): Please do not insult other forum members, or post rude comments.
SHOCK HORROR
dentafrices way worked :L Yours didnt xox
Joshh
06-04-2009, 08:21 PM
*Removed*
This is your code..
This checks to see if the rank is not 1, which I believe is the administrator rank.
If rank is not 1 (admin rank) Then display the message. You don't need the else, it exits.. so you don't need to else it.
if($logged["rank"] != "1") {
// not an admin.
exit("You need to be an administrator to view this feature"); // you can redirect to an error page, include something here, doesn't matter, just exit.
}
Now why *Removed* are you doing the same thing?
If rank is not admin rank Then display "You are an admin!" *Removed* You're saying they are an admin when they are not. Plus.. if they are not an admin.. it's not even going to get to this section anyway.
if($logged["rank"] != "admin rank here") {
// is an admin.
exit("You're an admin!"); // you can redirect to an error page, include something here, doesn't matter, just exit.
}
echo ("the contents of menus/anav.php here")
*Removed*
I see, I didn't think 1 was the administrator rank. And when I read it earlier I thought it said echo not exit sorry. :P
Dentafrice
06-04-2009, 08:23 PM
I see, I didn't think 1 was the administrator rank.
Well he indicated that when he posted it, that if it wasn't 1 then it wasn't an admin.
Joshh
06-04-2009, 08:24 PM
Well he indicated that when he posted it, that if it wasn't 1 then it wasn't an admin.
Yeah I obviously didn't read it propperly because when I read it earlier I also thought it said echo and not exit, just me not reading propperly. :P
Yeah I obviously didn't read it propperly because when I read it earlier I also thought it said echo and not exit, just me not reading propperly. :P
Lol so read it properly next time before you give advice ;) And no need to argue people ive fixed it now :) THanks alot to everyone :P
Dentafrice
07-04-2009, 01:18 PM
Lol so read it properly next time before you give advice ;) And no need to argue people ive fixed it now :) THanks alot to everyone :P
Uh oh, you better not say anything with curse words in it to him, you might get an infraction!
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