PHP Login Status Display

DarkLight posted this at 15:26 — 11th December 2007.

Joined: Oct 2007

Hi, I need a script that shows if you are logged.
if they are not, it says "Log In".
if they are, it says, "Log Out -username-"

Thanks in advance, this is very important.

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kazimmerman posted this at 20:12 — 11th December 2007.

He has: 698 posts

Joined: Jul 2005

Well, this would require the use of cookies and/or sessions, and here's how I would most simply do it:

A log-in form would process to this page:

if username and password are verified {
setcookie('giveItAName', $userNameVariable, $expiresWhen);
echo "You are now logged in.";
}

'
Then, on each page (I would place the code in a header that can be included on each page) you could see if the cookie exists:

if (isset($_COOKIE['giveItAName'])) {
$makeItAVariable = $_COOKIE['giveItAName'];
echo "Log out -" . $makeItAVariable . "-";
}
else {
echo "Log in";
}

'
You could make something much more elaborate than this, but this is basically how it works.

Kurtis

DarkLight posted this at 20:39 — 11th December 2007.

He has: 287 posts

Joined: Oct 2007

I cannot seem to get it to work, here is my "checklogin.php" script...

<?php
$host=\"localhost\";
$username=\"USERNAME\";
$password=\"PASSWORD\";
$db_name=\"DATABASE\";
$tbl_name=\"TABLE\";

// Connect to server and select databse.
mysql_connect(\"$host\", \"$username\", \"$password\")or die(\"cannot connect\");
mysql_select_db(\"$db_name\")or die(\"cannot select DB\");

// username and password sent from signup form
$username=$_POST['username'];
$password=$_POST['password'];

$sql=\"SELECT * FROM $tbl_name WHERE username='$username' and password='$password'\";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row

if($count==1){
// Register $myusername, $mypassword and redirect to file \"login_success.php\"
session_register(\"username\");
session_register(\"password\");
header('location:http://pcgenius.pcriot.com/members/basic/login/login_success.php');
}
else {
header('location:http://pcgenius.pcriot.com/members/basic/login/login_failed.php');
}
?>

What would I need to do to display the content of the created cookie on a different page?
Thanks you so much in advance, i am fed up of this now, i cannot seem to get the right script.

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kazimmerman posted this at 23:54 — 11th December 2007.

He has: 698 posts

Joined: Jul 2005

First of all, although I can't see it being the problem here, I would suggest creating variables with different names for the visitor's username and password and those for the MYSQL connection.

Also, I think you need the following line before you start registering values on a session:

session_start();

And where you try to do this, I think you should be inserting the variable, correct?

session_register('$username');
session_register('$password');

'
I'm not very experienced in using sessions, but this is how I would approach it based off of what I know.

Kurtis

DarkLight posted this at 22:26 — 12th December 2007.

He has: 287 posts

Joined: Oct 2007

AHHAH!!!
I sorted it.
all i did was add...

<?php
session_start();
?>

to every page i wanted the code to work on, it simply reminds the browser that it has a session in progress. thanks for your help. Very much appreciated, it was your idea that gave me inspiration to do this.

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