我无法让它工作，php 清理过滤器

有人能告诉我我做错了什么吗？我正在尝试使用 php 过滤器清理和验证表单用户输入。 此输入将转到 mySql 数据库，然后转到 html 页面输出，因此我需要删除所有 html 标记并为 mySQL 提供安全性。

谢谢。

      /*
 * BEGIN Sanitize and validate
 */
  // Adding fields to sanitize array.
$filters = array(
   'author'  => FILTER_SANITIZE_SPECIAL_CHARS,
   'title'  => FILTER_SANITIZE_SPECIAL_CHARS,
   'description'  => FILTER_SANITIZE_SPECIAL_CHARS,
);

/*** apply the filters to the POST array ***/
$filtered = filter_input_array(INPUT_POST, $filters);

 // filtering out anything that isn't an email address
    $sanitized_email = filter_var(($_POST["email"]), FILTER_SANITIZE_EMAIL);
    if ( filter_var($sanitized_email, FILTER_VALIDATE_EMAIL)  == TRUE) {
        echo '';
    } else {
        echo 'Invalid Email Address.  Go Back.';
        exit;
    }
    /*
 * BEGIN sanitize and validate
 */

新代码，但仍然无法正常工作。 我做错了什么。 仍然在数据库中获取html，并且错误不起作用。

      /*
 * BEGIN Sanitize and validate
 */
  // Adding fields to sanitize array.
$filters = array(
   'author'  => FILTER_SANITIZE_STRING,
   'title'  => FILTER_SANITIZE_STRING,
   'description'  => FILTER_SANITIZE_STRING,
);

/*** apply the filters to the POST array ***/
$filtered = filter_input_array(INPUT_POST, $filters);
$filters = array_map('mysql_real_escape_string',$filters);
$filters = array_map('htmlspecialchars',$filters);
    if ( filter_var_array($filtered, FILTER_SANITIZE_STRING)) {
        echo '';
    } else {
        echo 'Invalid Input.  Go Back.';
        exit;
    }

// filtering out anything that isn't an email address
    $sanitized_email = filter_var(($_POST['email']), FILTER_SANITIZE_EMAIL);
    if ( filter_var($sanitized_email, FILTER_VALIDATE_EMAIL)) {
        echo '';
    } else {
        echo 'Invalid Email Address.  Go Back.';
        exit;
    }
    /*
 * BEGIN sanitize and validate
 */

第2编如下：

通过将清理代码添加到将数据带到数据库的脚本区域，获取要在数据库中显示的清理数据。

 $_POST['author'] = mysql_real_escape_string(trim($_POST['author']));
      $author=filter_var($_POST['author'], FILTER_SANITIZE_STRING);
      $_POST['email'] = mysql_real_escape_string(trim($_POST['email']));
      $email=filter_var($_POST['email'], FILTER_SANITIZE_EMAIL);

现在我必须完成验证电子邮件的尝试。