user level issue with redirecting [duplicate] - php

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Closed 10 years ago.
Possible Duplicate:
checklogin condition issue in php
i have this quick question please,
i have this piece of code which isn't working properly, something about the syntax.. could you please help me with it?
i know it may sound stupid enough but i'm trying to understand!
Thanks!
<?php
session_start();
require_once('db.php');
include('functions.php');
if (checkLogin('1 2')) {
echo "hello ".$_SESSION['user_id']." You are now logged in.";
} else if (checkLogin('3')) {
echo "hey tst";
} else {}
?>
function checkLogin($levels)
{
if(!$_SESSION['logged_in'])
{
$access = FALSE;
}
else {
$kt = split(' ', $levels);
$query = mysql_query('SELECT Level_access FROM users WHERE ID = "'.mysql_real_escape_string($_SESSION['user_id']).'"');
$row = mysql_fetch_assoc($query);
$access = FALSE;
while(list($key,$val)=each($kt))
{
if($val==$row['Level_access'])
{//if the user level matches one of the allowed levels
$access = TRUE;
}
}
}
if($access==FALSE)
{
header("Location: login.php");
}
else {
//do nothing: continue
}
}
CREATE TABLE `users` (
`ID` int(11) NOT NULL auto_increment,
`Username` varchar(255) NOT NULL,
`Password` varchar(255) NOT NULL,
`Temp_pass` varchar(55) default NULL,
`Temp_pass_active` tinyint(1) NOT NULL default '0',
`Email` varchar(255) NOT NULL,
`Active` int(11) NOT NULL default '0',
`Level_access` int(11) NOT NULL default '2',
`Random_key` varchar(32) default NULL,
PRIMARY KEY (`ID`),
UNIQUE KEY `Username` (`Username`),
UNIQUE KEY `Email` (`Email`)
) ENGINE=MyISAM AUTO_INCREMENT=2 DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ;

Answer 1

Well you could simplify your checkLogin() function
function checkLogin($levels)
{
$access = false;
if (!isset($_SESSION['logged_in']) || !$_SESSION['logged_in'])
return false;
//use mysqli instead mysql
$con = new mysqli("localhost", "username", "password", "database");
$query = $con->query('SELECT Level_access FROM users WHERE ID = "'.$con->real_escape_string($_SESSION['user_id']).'"');
$row = $query->fetch_assoc();
$con->close();
if (in_array($row['Level_access'], explode(" ", $levels))) $access = true;
return $access;
}
This function should return true or false!
After that your code could look like this
session_start();
require_once('db.php');
include('functions.php');
if (checkLogin('1 2')) {
echo "hello ".$_SESSION['user_id']." You are now logged in.";
} else if (checkLogin('3')) {
echo "hey tst";
} else {
header("Location: login.php");
}
Hope this helps you.

Answer 2

Your if statements need parenthesis around them:
if( checkLogin('1 2')) {
^ ^

Answer 3

Try this
<?php
session_start();
require_once('db.php');
include('functions.php');
if (checkLogin('1 2')) {
echo "hello ".$_SESSION['user_id']." You are now logged in.";
} else if (checkLogin('3')) {
echo "hey tst";
} else {}
?>

Answer 4

Run the code in your browser. You'll get an error message. Use that error message to figure out what's wrong. Repeat until you get no error messages, and the program runs as designed.
That's how we debug things in the real world.