PHP - Logout User Not Working

this is not working

Code: [Select]
<?php

session_start();
$username = $_SESSION['username'];
echo mysql_query("UPDATE members SET online='0' WHERE username='$username' AND online='1'");
session_destroy();

echo "You've been logged out. <a href='index.php'>Click here</a> to return.";

?>
what im trying to do is change the online mysql data from 1 back to 0 as i want to show when a user is on and off line

can anyone help

Hi,

How logout is executed after 15 min. if does not hit or refresh any page.

Code: [Select]
<?php 
session_start();
$_SESSION['is_logged_out'] = 1;
if($_SESSION['is_logged_out'] ==1)
{
 echo("<SCRIPT LANGUAGE='JavaScript'>window.alert('You had been successfully signed out')</SCRIPT>");
 session_unset();
 session_destroy();
 header("Location:chatframe.php");
}
?>
I got this error   Warning: Cannot modify header information - headers already sent by (output started at C:\xampp\htdocs\login\logout.php:6) in C:\xampp\htdocs\login\logout.php on line 9


which mistake im put?

Hey guys, me again. A few of you were helping me with my login script, which I did finally get working. I am having one small problem though. Upon clicking the logout link, it does not re-direct back to the login page, it just stays blank. I have pasted my code from my logout.php file for reference. Thanks in advance for your help!

Logout.php code

<?php
session_start();
session_destroy(); {
header("location: login.php");
}
exit();
?>

Hi, im new here 

Need some help.

I have a database which contains users, and for every user i have logged_in value which is 1 if user is logged in and 0 if not. I use this so that user can be logged in only once.
When user clicks on logout link i set logged_in on 0.
My problem is when browser closes logged_in for current user is not set to 0, and after that user can't login because my site thinks he is already logged in.

is there some way I can call some function when browser exits or any other solution?

please help

When writing a logout script in PHP what steps should i take? Should i:

1.) unset all session vars like:  

 unset($_SESSION["var1"]);
 unset($_SESSION["var2"]);

or calls   

session_unset();

2.) destroy session cookie like:

setcookie(session_name(), "NULL", time()-60, $params['path'], $params['domain'], $params['secure'], isset($params['httponly']));

3.) finally destroy the session data by calling

session_destroy();

or maybe just only unset the session vars - that is making only the first step?

I'm asking because I've seen many logout scripts that doesn't involve these 3 steps (i most cases doing only the first one as i stated)

Thx. for help.

how to logout using session in proper way ? can you give me the code

I had set this value into the super global $_SESSION in the first file, file_1.php:

<?php

//there is some code here

$_SESSION['salesorder'] = 'are---io';

//remaining code, includes a form.

?>

When the submit button is clicked, the form is submitted to file_2.php 

<?php
//some code over here
if (isset($_POST['sales']) && $_POST['sales'] != ""){
    $sales = sanitize_input(trim($_POST['sales']));	
    $_SESSION['salesorder'] = $sales;
	$host  = $_SERVER['HTTP_HOST'];
	$uri   = rtrim(dirname($_SERVER['PHP_SELF']), '/\\');
	$extra = 'processSO_helper_sd.php';
	header("Location: http://$host$uri/$extra?so=".$sales);
	exit();
 }
//Begin display
 include 'inc_fn_header_and_menu.php';
//some code over here
?>

which will redirect to processSO_helper_sd.php

<?php

// Initialize session
session_start();

    $SD_ID = $_SESSION['salesorder'];
    $result = $db->query($sql);
	$row = $result->fetch_assoc();
	$rowcount =$row['row_count'];*/
	//print_r($_SESSION);//*
	$result = $tmonedb->query("SELECT COUNT(*) FROM document WHERE SD_ID = $SD_ID")->fetch_array();
	$rowcount = $result[0];
//remaining code

?>

My problem is that I cannot access the 

$_SESSION['salesorder'] in the last file processSO_helper_sd.php although I have set it twice previously. 

I could not add session_start(); in the file_2.php  because that was already added in  'inc_fn_header_and_menu.php'; 

Can someone help me with this? Thank you. Please tell me if you need any additional information.

I was wondering if there is away to set a logout message when you are using sessions

this is my code

<?php
     session_start();
      if($_SESSION[auth] = "yes") {
      session_unset(); 
      session_destroy();
       header( "Location: index.php" );
      exit();
     } else { 
       if ($_SESSION[auth] != "yes") {
       header( "Location: index.php" );

   $_SESSION['error'] = "please login!";
      exit();
    }
  }
?>

Hi,

I am having a bit of problem with my login/logout script. When user is logged in I want the script to show logout and if they are not login I want the script to show login. The problem is even when the user is logged in it says "you must be logged in Click here to login " here is the script Please help

Code: [Select]


<?php
session_start();
$_SESSION['username'] = $_POST['username'];
if ($_SESSION['username'])
echo "Welcome, ".$_SESSION['username']."!<br><a href='logout.php'>Logout</a>" ;
else
 die("you must be logged in <a href='Login.php'>Click here to login</a>");
?>


what did I do wrong in this script ?
Thanks

is it possible to update data in mysql when a user clicks logout?

Hi there I have this problem with a button that should destroy sessions and then redirects to login page when clicked, pointing to this method:

Code: [Select]
public function logout(){
$this->logged_in = false;
session_start();
session_unset();
session_destroy();
redirect_to("login.php");
    }
the function is called as bellow:
Code: [Select]
             <form>
    <input id="logout" name="logout" type="submit" value="logout" onClick="<?php $session->logout(); 
    ?> " />
</form>

which is available on a page called index.php, which is the default page where you'll be directed to when logged in or just registered. unfortunately when I'm supposed to be to this page and before even clicking on the button, the function just does the work and redirects to the login page!!!.....any help please







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Hey!

I have a table...with a column isOnline which is set to 1 when logging in (using login.php)...and 0 when logging out (logout.php), its how I determine whos logged in and whos not (in statistics etc.)

Now the problem is if they dont have the remember me enabled..and don't log themselfs out via logout.php (which sets the isOnline to 0)...and close their browser the $_SESSIONS's get destroyed (which means their physically not logged in)...and their isOnline remains at 1 (even though their logged out)

Another problem..is if they login with remember me enabled (as they want to remain logged in)....and close their browser or whatever, isOnline will still remain as 1 (theirfore others will think their online).

For your info:
In login.php I do an UPDATE query adding the current time generated by time() into the loginTime column...and also setting isOnline to 1.

So my question is how can I overcome this, I have the isOnline (1 = logged in, 0 not logged in) and loginTime (which contains the login timestamp)?

Cheers

I based this off some other pages read, and think I'm doing this wrong or it's just not connecting. 

Code: [Select]
Here's the database table:
CREATE TABLE users (
    id INT NOT NULL AUTO_INCREMENT,
    username VARCHAR(30) NOT NULL UNIQUE,
    password VARCHAR(64) NOT NULL,
    salt VARCHAR(3) NOT NULL,
    PRIMARY KEY(id)
);


Ando far I have index.php with my login form <form name="login" action="login.php" method="post">
    Username: <input type="text" name="username" />
    Password: <input type="password" name="password" />
    <input type="submit" value="Login" />
</form><br />Would you like to <a href="register.php">register?</a></center>

Then I have my actual login on login.php (header.php includes website's main image as well as session_start():
<?php

include('header.php');

$username = $_POST['username'];
$password = $_POST['password'];
//connect to the database here
$username = mysql_real_escape_string($username);
$query = "SELECT password, salt
        FROM users
        WHERE username = '$username';";
$result = mysql_query($query);
if(mysql_num_rows($result) < 1) //no such user exists
{
    header('Location: login.php');
    die();
}
$userData = mysql_fetch_array($result, MYSQL_ASSOC);
$hash = hash('sha256', $userData['salt'] . hash('sha256', $password) );
if($hash != $userData['password']) //incorrect password
{
    header('Location: login_form.php');
    die();
}
else
{
    validateUser(); //sets the session data for this user
}
//redirect to another page or display "login success" message
?>

then I have my register php on register.php:
<?php

include('header.php');

//retrieve our data from POST
$username = $_POST['username'];
$pass1 = $_POST['pass1'];
$pass2 = $_POST['pass2'];
if($pass1 != $pass2)
    header('Location: register_form.php');
if(strlen($username) > 30)
    header('Location: register_form.php');
    
    $hash = hash('sha256', $pass1);
    function createSalt()
{
    $string = md5(uniqid(rand(), true));
    return substr($string, 0, 3);
}
$salt = createSalt();
$hash = hash('sha256', $salt . $hash);

$dbhost = 'localhost';
$dbname = 'mygame';
$dbuser = 'root';
$dbpass = '';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db($dbname, $conn);
//sanitize username
$username = mysql_real_escape_string($username);
$query = "INSERT INTO users ( username, password, salt )
        VALUES ( '$username' , '$hash' , '$salt' );";
mysql_query($query);
mysql_close();
header('Location: login.php');

?>
and lastly the register form:
   
    <center><form name="register" action="register.php" method="post">
    Username: <input type="text" name="username" maxlength="30" />
    Password: <input type="password" name="pass1" />
    Password Again: <input type="password" name="pass2" />
    <input type="submit" value="Register" />
</form></center>
I am getting the errors:
Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in /Applications/XAMPP/xamppfiles/htdocs/testing/login.php on line 13

Warning: Cannot modify header information - headers already sent by (output started at /Applications/XAMPP/xamppfiles/htdocs/testing/config.php:1) in /Applications/XAMPP/xamppfiles/htdocs/testing/login.php on line 15

Could someone explain why this is happening?

Hi,

Just wondered if anyone could help 

Ive been following this tutorial: http://net.tutsplus.com/tutorials/php/user-membership-with-php/

Ive got a simple membership system working now, but just wondering about the login / login links that i currently have. The login link is currently hard coded like so:

Code: [Select]
<ul id="menu">
<li id="active"><a href="index.html">Home</a></li>
<li><a href="About.html">About</a></li>
<li><a href="Contact.php">Contact</a></li>
<li class="end"><a href="login.php">Login</a></li>
</ul>

and same for the logout:
Code: [Select]
<ul id="menu">
<li id="active"><a href="index.html">Home</a></li>
<li><a href="About.html">About</a></li>
<li><a href="Contact.php">Contact</a></li>
<li class="end"><a href="logout.php">Logout</a></li>
</ul>
But the problem is, when i go to the about us page for example it will still display the login which really it should have logout.

Could anyone offer some assistance please