Hi Every One 
I got one form … each category has different ID…
Example:-
http://localhost/control/newdpppage/dpp_add_lorthbank.php?cat=3
…
http://localhost/control/newdpppage/dpp_add_lorthbank.php?cat=4
Each user after login they go to the page assigned for them …
-The Question…
How do i prevent User1 that can see cat=3 to see the data on cat=4 assigned for User2.
Thank you 
I use for that JavaScript
<SCRIPT language=JavaScript>
function reload(form)
{
var val=form.cat.options[form.cat.options.selectedIndex].value;
self.location='dpp_add_lorthbank.php?cat=' + val;
}
</script>
Hi,
you can check the database on the top of the dpp_add_lorthbank.php page and redirect to a customized page if the user do not have permission to the category.
suppose a user with uid 5 have permission to category with catid 2 and your database table structure is like:
uid catid
1 3
1 2
5 2
5 1
now you need to check if a user requesting the page matches the catid and user id or not.
example (Assuming the relationship table as usercategory and user id is in session):
assuming that you have stored userid in session.
$sql=mysql_query("select * from usercategory where uid=".$_SESSION['uid']." AND catid=".$_GET['cat']."");
if(mysql_num_rows($sql)){
//let the user do stuffs
}else{
//redirect to error page or die("you dont have permission to this category")
}
Thank you
bishwadeepkc
I tried what you told me, i might i’ve problems make it the right way…
But i tried and this is was the result … 
Please explain for me more.
<?php
//Connect to mysql server
$link = mysql_connect('localhost', 'root', '');
mysql_select_db("sac", $link);
if(!$link) {
die('Failed to connect to server: ' . mysql_error());
}
$sql=mysql_query
("select * from login_type, category where
uid=".$_SESSION['uid']." AND cat_id=".$_GET['cat_id']."");
if(mysql_num_rows($sql)){
//let the user do stuffs
echo "<a href='logout.php'>Logout</a>";
}
else
{
//redirect to error page or die("you dont have permission to this category")
die(mysql_error());
}
?>
First Error:-
Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\control
ewdpppage\dpp_add_lorthbank.php on line 26
Second Error:-
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ‘[0] AND cat_id=’ at line 2
SELECT *
FROM login_type, category
WHERE uid =2
AND cat_id =2
This Sql Statement worked for me … I’ll try how’s it gonna be with the first error which still i cant find the solution :injured:
Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\control
ewdpppage\dpp_add_lorthbank.php on line 26
You are joining two tables without defining the join criteria, so the result would be a cross join. I’m guessing that cat_id is present in both tables? That would give you a mysql error, and the result would not be a mysql result set (hence the error you’re seeing). Try this:
$sql = "
SELECT *
FROM login_type
WHERE uid = " . $_SESSION['uid'] . "
AND cat_id = " . $_GET['cat_id']
;
$result = mysql_query($sql) or die('mysql error ' . mysql_error() . ' in query ' . $sql);
if(mysql_num_rows($result) > 1){
//let the user do stuffs
echo "<a href='logout.php'>Logout</a>";
}
else
{
//redirect to error page or die("you dont have permission to this category")
die(mysql_error());
}
By the way, don’t use user input ($_GET, $_POST) in your queries without sanitizing the data first (in this case, make sure $_GET[‘cat_id’] contains a numeric value).
You are joining two tables without defining the join criteria, so the result would be a cross join. I’m guessing that cat_id is present in both tables?
I’ve two tables
login_type:-
uid| type | username | password
1 | adminbank | admin | 123
2 | userbank | mark | 321
category:-
cat_id | category
1 | EonBank
2 | MayBank
I’ve one form shows the categories by this way dpp_add_lorthbank.php?cat=1 for user one.
cat=2 for user two.
what i’m trying to do is making each one only see his category.
…
user admin after login go to the link *.php?cat=1
user mark after login go to the link *.php?cat=1
both cant see the values of the category for each.
How do you know what category to show to what user? What is the connection between the two tables?
uid catid
1 3
1 2
5 2
5 1
You posted this. Where do you get that info from?
That was suggestion from bishwadeepkc
I tried to do like what he suggest but couldn’t get it work …
what i post up was my original data.
Thank you
Mr.guido2004
I tried to do like this
just an idea
I’m asking is that would help in my case?
First page
<?php
session_start();
$key = 14;
$_SESSION['cat'] = $key;
echo "<a href=\\"second.php?cat=$key\\">link</a>";
?>
Second Page
<?php
session_start();
if($_GET['cat'] != $_SESSION['cat'])
{
echo "You are NOT the Same Person from Login";
exit;
}
session_destroy();
?>
Yes, you can store the user’s category id in a session. You don’t even have to pass the category id in the query string anymore. Just get it from the session.
On the login page i did this
<?php
//Start session
session_start();
$_SESSION['foo']='$_POST[username]';
$_SESSION['uid']='$usercategory';
$key = 14;
$_SESSION['cat'] = $key;
?>
Second page
if ($_GET['cat'] != $_SESSION['cat'])
//echo "You are NOT the Same Person from Login";
header('Location: error.php');
else
echo "You are THE Same Person from Login";
session_destroy();
?>
And i added the sql statment…
All working well now …
no mistakes at all… 
Thank you
guido2004
and
bishwadeepkc
:argue:
The last code only shows the record of cat=14 because it’s $key = 14;
How i can make it
$key = 14,15,16;
I tried
$key = array (14,15,16);
also
$key14 = 14;
$key15 = 15;
$key16 = 16;
But all wrong 