在这个网站的帮助下,我现在已经让我的登录脚本在很大程度上工作了,但是当我试图检查一个受限制的页面上的会话时,有一个问题,完整的代码如下。
登录脚本本身
<?php
function validateUser()
{
session_regenerate_id (); //this is a security measure
$_SESSION['valid'] = 1;
$_SESSION['userid'] = $userid;
}
?>
<?php
ob_start(); // Start output buffering
error_reporting(E_ALL & ~E_NOTICE);
ini_set('display_errors', TRUE);
ini_set('display_startup_errors', TRUE);
session_start(); //must call session_start before using any $_SESSION variables
$username = $_POST['username'];
$password = $_POST['password'];
//connect to the database here
$hostname_PropSuite = "localhost";
$database_PropSuite = "propsuite";
$username_PropSuite = "root";
$password_PropSuite = "root";
$PropSuite = mysql_pconnect($hostname_PropSuite, $username_PropSuite, $password_PropSuite) or trigger_error(mysql_error(),E_USER_ERROR);
mysql_select_db($database_PropSuite, $PropSuite);
$username = mysql_real_escape_string($username);
$query = "SELECT password, salt FROM admin_users WHERE username = '$username';";
$result = mysql_query($query) or die(mysql_error());
if(mysql_num_rows($result) < 1) //no such user exists
{
header('Location: http://localhost/PropSuite/index.php?login=fail');
die();
}
$userData = mysql_fetch_array($result, MYSQL_ASSOC);
$hash = hash('sha256', $userData['salt'] . hash('sha256', $password) );
if($hash != $userData['password']) //incorrect password
{
header('Location: http://localhost/PropSuite/index.php?login=fail');
die();
}
else
{
validateUser(); //sets the session data for this user
}
//redirect to another page or display "login success" message
header('Location: http://localhost/PropSuite/main');
die()
//redirect to another page or display "login success" message
?>我试图限制的页面。
<?php
error_reporting(E_ALL & ~E_NOTICE);
ini_set('display_errors', TRUE);
ini_set('display_startup_errors', TRUE);
function isLoggedIn()
{
if(isset($_SESSION['valid']) && $_SESSION['valid'])
return true;
return false;
}
//if the user has not logged in
if(!isLoggedIn())
{
header('Location: http://localhost/PropSuite/index.php');
die();
}
//page content follows
?>当我按下log时,它会把我带回登录页面,所以它看起来像是通过登录脚本,然后当它到达主页时,它会把我带回登录页面。我是confused.com :-(
预先感谢您的帮助
发布于 2012-10-24 21:44:37
在您试图限制的页面中,我看不到session_start
<?php
session_start();
error_reporting(E_ALL & ~E_NOTICE);
ini_set('display_errors', TRUE);
ini_set('display_startup_errors', TRUE);
.....
?>顺便说一句,在你的其他脚本中,你应该这样写
<?php
.....
$username = isset($_POST['username'])?$_POST['username']:'';
$password = isset($_POST['password'])?$_POST['password']:'';
...
?>https://stackoverflow.com/questions/13050460
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