Wolverine68 Posted March 7, 2009 Share Posted March 7, 2009 I posted a sessions topic earlier today where I had the user login with a name and password. If successful it created an active session and displayed the appropriate message. If not, it displayed an "access denied" message. Now, I want to add a logout feature. If the user is successful in logging into a session, there will be a button they click on when they're ready to logout. After clicking the button, it takes them to a logout.php page. The session is destroyed and a message displayed saying they have logged out. Login.php page: <?php session_start(); $_SESSION['name'] = "ken"; $_SESSION['password'] = "welcome"; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html> <head> <title>Session Variables</title> </head> <body> <div> <form action="display.php" method="POST"> <p>Enter your username and password to login:</p> <p>Name:<INPUT TYPE="text" SIZE="20" name="name"></p> <p>Password:<INPUT TYPE="text" SIZE="20" name="password"></p> <p><INPUT TYPE="submit" VALUE="Submit!"></p> </form> </div> </body> </html> Display.php page: <?php session_start(); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html> <head> <title>Validating Session</title> </head> <body> <div> <?php if (($_POST['name'] == $_SESSION['name']) && ($_POST['password'] == $_SESSION['password'])) { $_SESSION['Active'] = "True"; } else { $_SESSION['Active'] = "False"; } ?> <?php If ($_SESSION['Active'] == "True") { echo "Welcome, " .$_POST['name']. "you have successfully logged into your session."; } else { echo "You have entered an invalid username and/or password. Access denied"; } ?> <form action="logout.php" method="POST"> <p>Click on the following button when you want to logout of your session: <input type="submit" value="Logout"></p> </form> </div> </body> </html> Logout.php page: <?php session_start(); unset($_SESSION['Active']); session_destroy(); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html> <head> <title>Session Variables</title> </head> <body> <div> <p><?php echo "You have logged out of your session"; ?> </div> </body> </html> Even if the user entered a wrong name or password, they're still going to see the message "Click on the button to logout". Of course, I only want that message to appear if the login was successful, since they can't logout if they weren't successful in logging in to begin with. Quote Link to comment Share on other sites More sharing options...
wildteen88 Posted March 7, 2009 Share Posted March 7, 2009 Change <?php If ($_SESSION['Active'] == "True") { echo "Welcome, " .$_POST['name']. "you have successfully logged into your session."; } else { echo "You have entered an invalid username and/or password. Access denied"; } ?> <form action="logout.php" method="POST"> <p>Click on the following button when you want to logout of your session: <input type="submit" value="Logout"></p> </form> to <?php If ($_SESSION['Active'] == "True") { echo "Welcome, " .$_POST['name']. "you have successfully logged into your session."; echo '<form action="logout.php" method="POST"> <p>Click on the following button when you want to logout of your session: <input type="submit" value="Logout"></p> </form>';' } else { echo "You have entered an invalid username and/or password. Access denied"; } ?> Quote Link to comment Share on other sites More sharing options...
Wolverine68 Posted March 7, 2009 Author Share Posted March 7, 2009 Thank you. Quote Link to comment Share on other sites More sharing options...
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