PHP Online users--- Sessions function - php

Question

When a person logs in , the online column in the table is set to 1 , and when he logs out , it is set to 0 . I achieved the Login script , but problems on ending the session with SQL query .. Please do help !
There is No Error Displayed but Online value remains as 1 even after logging out
**LOGOUT SCRIPT**
<?php
$offline = $_SESSION["username"] ;
?>
<?php
//If the user is logged, we log him out
if(isset($offline))
{
//We log him out by deleting the username and userid sessions
unset($_SESSION['username'], $_SESSION['userid']);
$con=mysqli_connect("localhost","root","","chat");
mysqli_query($con,"UPDATE users SET Online=0
WHERE username='.$offline.'");
mysqli_close($con);
?>
LOGIN SCRIPT
<?php
$ousername = '';
//We check if the form has been sent
if(isset($_POST['username'], $_POST['password']))
{
//We remove slashes depending on the configuration
if(get_magic_quotes_gpc())
{
$ousername = stripslashes($_POST['username']);
$username = mysql_real_escape_string(stripslashes($_POST['username']));
$password = stripslashes($_POST['password']);
}
else
{
$username = mysql_real_escape_string($_POST['username']);
$password = $_POST['password'];
}
//We get the password of the user
$req = mysql_query('select password,id from users where username="'.$username.'"');
$dn = mysql_fetch_array($req);
//We compare the submited password and the real one, and we check if the user exists
if($dn['password']==$password and mysql_num_rows($req)>0)
{
//If the password is good, we dont show the form
$form = false;
//We save the user name in the session username and the user Id in the session userid
$_SESSION['username'] = $_POST['username'];
$_SESSION['userid'] = $dn['id'];
$con=mysqli_connect("localhost","root","","chat");
$sql = mysql_query('UPDATE users SET Online=1 where username="'.$username.'"');
?>
<div class="message">You have successfuly been logged. You can access to your member area.<br />
Home</div>
<?php
}
else
{
//Otherwise, we say the password is incorrect.
$form = true;
$message = 'The username or password is incorrect.';
}
}
else
{
$form = true;
}
if($form)
{
//We display a message if necessary
if(isset($message))
{
echo '<div class="message">'.$message.'</div>';
}
//We display the form
?>
<div class="content">
<form action="connexion.php" method="post">
Please type your IDs to log in:<br />
<div class="center">
<label for="username">Username</label><input type="text" name="username" id="username" value="<?php echo htmlentities($ousername, ENT_QUOTES, 'UTF-8'); ?>" /><br />
<label for="password">Password</label><input type="password" name="password" id="password" /><br />
<input type="submit" value="Log in" />
</div>
</form>
</div>
And , when I do the SQL query , How do I do the if Online = 1 , display online.png ? else 'Blank space ' ?
Thanks in advance !

Answer 1

The below code is for logout.php
<?php
//If the user is logged, we log him out
if(isset($_SESSION['username']))
{echo $_SESSION['username'];
//We log him out by deleting the username and userid sessions
$username=$_SESSION['username'];
$sql="UPDATE users SET online=0 WHERE username='$username'";
mysql_query($sql);
unset($_SESSION['username'], $_SESSION['userid']);
?>
And this one for sign_up.php
if(isset($_POST['username'], $_POST['password'], $_POST['passverif'], $_POST['email'], $_POST['username']))

Answer 2

use session_start();
ANd session_destroy(); FOR LOGOUT.