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>
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