1

可能重复：
如何在 PHP 中实现类似 SO 的 returnurl？

我正在尝试这样做，以便当未登录的用户浏览我的网站并单击链接时，他们会被带到登录页面（在 jQuery 灯箱窗口中打开），登录后他们会被带回他们所在的上一个/原始页面。

我已经尝试使用这个简单的代码来尝试实现这一点，但是当我使用它时，它不会显示登录页面并直接进入上一个/原始页面，因此用户无法登录。

if(!empty($_SERVER['HTTP_REFERER']))
{
  header('Location: '.$_SERVER['HTTP_REFERER']);
}

这是我所有的登录脚本代码：

HTML：

<form action="login.php" method="post" target="_top"  >
  <div class="row email">
    <input type="email" id="email" name="email" placeholder="Email" value="<?php echo htmlentities($email); ?>" />
  </div>
  <div class="row password">
    <input type="password" id="password" name="password" placeholder="Password" value="<?php echo htmlentities($email); ?>" />
  </div>
  <input type="submit" name="submit" class="submit-login" value="Login >
</form>

PHP：

<?php
  if (logged_in())
  {
    redirect_to("dashboard.php");
  }

  include_once("includes/form_functions.php");

  // START FORM PROCESSING
  if (isset($_POST['submit'])) { // Form has been submitted.
    $errors = array();

    // perform validations on the form data
    $required_fields = array('email', 'password');
    $errors = array_merge($errors, check_required_fields($required_fields, $_POST));

    $fields_with_lengths = array('email' => 30, 'password' => 30);
    $errors = array_merge($errors, check_max_field_lengths($fields_with_lengths, $_POST));

    $email = trim(mysql_prep($_POST['email']));
    $password = trim(mysql_prep($_POST['password']));
    $hashed_password = md5($password);

    if ( empty($errors) ) {
      // Check database to see if email and the hashed password exist there.
      $query = "SELECT id, email, close_account ";
      $query .= "FROM ptb_users ";
      $query .= "WHERE email = '{$email}' ";
      $query .= "AND password = '{$hashed_password}' ";
      $query .= "AND close_account = '0' ";
      $query .= "LIMIT 1";
      $result_set = mysql_query($query);
      confirm_query($result_set);
      if (mysql_num_rows($result_set) == 1) {
        // email/password authenticated
        // and only 1 match
        $found_user = mysql_fetch_array($result_set);
        $_SESSION['user_id'] = $found_user['id'];
        $_SESSION['email'] = $found_user['email'];
        $_SESSION['sub_expires'] = $found_user['subscription_expires'];

        $result = mysql_query("UPDATE ptb_users SET user_online='Online' WHERE id=".$_SESSION['user_id']."") 
                or die(mysql_error());


        redirect_to("dashboard.php");
      } else {
        // email/password combo was not found in the database
        $message = "<div class=\"infobox\"><strong>Email/Password combination incorrect.</strong><br />"
                .= "Please make sure your caps lock key is off and try again.</div><br/>";
      }
    } else {
      if (count($errors) == 1) {
        $message = "<div class=\"infobox\">There was 1 error in the form.<div>";
      } else {
        $message = "<div class=\"infobox\">There were " . count($errors) . " errors in the form.<div>";
      }
    }
  } else { // Form has not been submitted.
    if (isset($_GET['logout']) && $_GET['logout'] == 1) {
      $message = "<div class=\"infobox\">You are now logged out.</div>";
    } 
    $email = "";
    $password = "";
  }
?>
<?php if (!empty($message)) {echo "<p class=\"message\">" . $message . "</p>";} ?>
<?php if (!empty($errors)) { display_errors($errors); } ?>
4

2 回答 2

2

当有人尝试访问需要身份验证的页面时，将该 URL 存储在 $_SESSION 变量中。然后将它们重定向到登录页面。登录后检查 $_SESSION 变量是否存在。如果是这样，请将它们重定向回该 URL。

于 2013-01-22T14:06:38.953 回答
1

您可以在 AJAX 的帮助下登录，成功登录后重新加载页面。这在很多方面都会更有帮助。

例如，如果用户名/密码错误，则查询灯箱窗口中可能会显示错误，而不是等待页面加载显示错误。

于 2013-01-22T14:10:31.007 回答