php检查用户否等已经登录的代码,通过Session判断
functions.php <?php function loggedIn(){ //Session logged is set if the user is logged in //set it on 1 if the user has successfully logged in //if it wasn't set create a login form if(!$_SESSION['loggd']){ echo'<form action="checkLogin.php" method="post"> <p> Username:<br> <input type="text" name="username"> </p> <p> Password:<br> <input type="password" name="username"> </p> <p> <input type="submit" name="submit" value="Log In"> </p> </form>'; //if session is equal to 1, display //Welcome, and whaterver their user name is }else{ echo 'Welcome, '.$_SESSION['username']; } } ?> index.php <?php //Start the session session_start(); //This is a simplified HTML Document ?> <html> <head> <title>My Page</title> </head> <body> <?php //Call the functions file require_once("functions.php"); //Display either the user's name, or the login form //This can be placed on many pages without having //to re-write the form everytime, just use this function logedIn(); ?> </body> </html>