ID: 114
Viewed: 11397
Added: Apr 08, 2002
Version:
Snippet uploaded by: snippet
Written By: snippet
Demo: Sorry, no demo
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A basic cookie script that writes a cookie and deletes a cookie based on the user who logs in or out. It checks against a mysql database to ensure the user actually exist.
<!---Head--->
none<!---Body--->
<?
$link = mysql_connect("localhost", "root", "*****");
mysql_select_db("dbname");
IF ($action == "logout"){
// actually deletes teh cookie if one is set.
setcookie("cookie_name", "", time() - 43200);
echo "Cookie has been deleted.";
exit;
}
IF ($action == "login")
{
//check use\r\name and password against a mysql database.
$result = mysql_query("SELECT * FROM Users WHERE name='$uname' AND pass='$upassword'");
$check = mysql_num_rows($result);
//if not row was found then write failed
IF (!$check){
echo "Failed to log in, sorry";
exit;
}
else{
//row was found so it writes the cookie and lets the user on.
setcookie ("cookie_name", "$u_name", time() + 43200);
echo "Welcome $u_name, you <br>";
echo "loggen in ok";
//redirects the user to another page once logged in.
echo"<META HTTP-EQUIV="Refresh" Content="1;URL=index.php">";
}
mysql_close($link);
}else{
?>
<html>
<head></head>
<body>
<form action="<?= $PHP_SELF ?>" method="post">
Name<input type=text name=u_name>
Pass<input type=password name=u_password>
<input type=hidden name=action value=login>
<input type=submit value="send">
</form>
<form action="<?= $PHP_SELF ?>" method="post">
<input type=hidden name=action value=logout>
<input type=submit value="Log Out">
</form>
</body>
</html>
<?
}
?>
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