如何仅为已登录用户保护页面？

Question

我创建了一个效果很好的登录表单.但我意识到我的用户所指向的页面仍然可以被任何人访问.如何保护只有登录用户可以查看的被访问页面？

我是否需要在成功页面上放置脚本？

这是我的check_login.php:

<?php
$host="localhost"; // Host name
$username="xxx"; // Mysql username
$password="xxx"; // Mysql password
$db_name="xxx"; // Database name
$tbl_name="xxx"; // Table name

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

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

// To protect MySQL injection (more detail about MySQL injection)
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$myusername = mysql_real_escape_string($myusername);
$mypassword = mysql_real_escape_string($mypassword);

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

$count=mysql_num_rows($result);

$user_info = mysql_fetch_assoc($result);

 if( isset($user_info['url']) ) {

 session_register("myusername");
 session_register("mypassword");
     header('Location: ' . $user_info['url']); //Redirects to the supplied url from the DB
 } else {
  header("location:error.htm");
 }
 ?>

Answer 1

每个页面都应该以

session_start();

并且您不应该使用session_register( "variablename" )PHP 4.2版本,使用

$_SESSION["variable"] = value;

所以带有is-logged-it检查的示例页面将是:

<?php
session_start();
if($_SESSION["loggedIn"] != true) {
    echo("Access denied!");
    exit();
}
echo("Enter my lord!");
?>

和登录脚本:

<?php
    /*
        ... db stuff ...
    */

if( isset($user_info['url']) ) {
    $_SESSION["loggedIn"] = true;
    $_SESSION["username"] = $myusername;
    header('Location: ' . $user_info['url']); //Redirects to the supplied url from the DB
} else {
    header("Location: error.htm");
}
?>