[RESOLVED] Coding User_Level - Problems

[Lozmax]

Hello, i am trying to fix so when i login it will run a sql check for admin status,
i have admin row in the database, set to tinyint, and default is set to 0.
Which is user, and 1 for admin, and then when the sql has been made
set the results into a session.

$sqln = "SELECT *
	FROM user
	WHERE admin = 0
	"; 
$rs = mysql_query($sqln) or die(mysql_error());
if (mysql_num_rows($rs) < 0) {
$_SESSION = mysql_fetch_array($rs);
}

no error is given at this point.

What is the use for this?
A: well i made a pm script and i want
to be able to if admin session is detected for that user.
That she/he will be able to choose from mass pm form or regular form.
Or if it's only user that is logged in to be able to only send pm to one user
at the time.

Thank you in advance.
-Loz

[djjjozsi]

Expecting more then one row means you cannot fetch all the results in a single mysql_fetch_array() row. What do you need here?
Query the user details,
based on his/her user inputs (username / password)
Lets create a query based on these values: $POST["username"] and $POST["password"]

And if there is a match in the users table, save the userlevel field in a session variable, to be able to check his/her user level in your PM system.

How could it be true?

mysql_num_rows($rs) < 0

- less then 0? But mysql_num_rows returns 0 or greater then 0 based on the number of records.

[halojoy]

What you need for find the row of a special user
is either 'user_name' or 'user_id' (or current user's unique email).

This is something you get when a user login.
When $_SESSION['user_id'] is set from database.

Something like this is what you should use.

<?php

if(isset($_SESSION['user_id'])){//only if is logged in

$sess_uid = $_SESSION['user_id'];

//Test if user_id is ADMIN
$sqln = "SELECT admin
FROM user
WHERE user_id = $sess_uid
";
$rs = mysql_query($sqln) or die(mysql_error());

if(mysql_num_rows($rs)>0){//found user_id?

    $row = mysql_fetch_array($rs);
    if($row['admin'] == 1)
        $is_adm = true;
    else
        $is_adm = false;

    // display result as a test
    echo 'User is '.($is_adm? 'Admin':'not admin');

}else{
	echo 'User id: '.$sess_uid.' not in db!!!';
} 

}

exit();
?>

[Lozmax]

thx, halojoy.
i changed the code a little to work for my code
as when i set a user i only save the username and password
in session at the moment. And all is in $SESSION[valid_user],
but something i added is not right i know it in my bones but,
i am not getting it at the momment (tired)..:p

$ss = $_SESSION[valid_user];
    $id = $sess_uid;


$sqln = "SELECT admin
FROM user
WHERE id = $sess_uid
";
$rs = mysql_query($sqln) or die(mysql_error());

if(mysql_num_rows($rs)>0){//found user

    $row = mysql_fetch_array($rs);
    if($row['admin'] == 1)
        $is_adm = true;
    else
        $is_adm = false;

    // display result as a test
    echo 'User is '.($is_adm? 'Admin':'not admin');

}else{
    echo 'User id: '.$sess_uid.' not in db!!!';
}

[Lozmax]

if(isset($_SESSION[valid_user)){
forgot to add that in the posted code.

[djjjozsi]

Where did you start session?

session_start();

Before you run a query, echo its contents:

echo "The query is: $sqln";

If you don't know what is in the $SESSION array,
print all indexes, and start troubleshooting:

print_r($_SESSION);