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我正在尝试编写一个执行以下操作的脚本:将用户消息打印到屏幕并将其通过电子邮件发送给他们。

这是我的代码 出于某种原因,当它是有效的电子邮件地址时,它告诉我我的电子邮件地址无效,有什么建议吗?我错过了什么?:-\

<?php
  if(strtolower($_SERVER['REQUEST_METHOD']) == 'post');

  $errors = null;
  $success = true;

  function checkEmail($email){
      return filter_var($email, FILTER_VALIDATE_EMAIL);
}

 if($_POST){
$errors = array();
$to = $_POST['email'];
$subject = 'Your Comment';
$message = $_POST['message'];
$headers = 'From: blahblah69@gmail.com' . "\r\n" . 'Reply-To:
     mitides.constantin@gmail.com' . "\r\n" .
            'X-Mailer: PHP/' .phpversion();



if(!$to || !$message){
    $errors[] ="Please fill in both a email and a message.";
} else if(!checkEmail($to) && $_POST['message'] = filter_var($_POST['message'],
   FILTER_SANITIZE_STRING));{
    $errors[]= print ('You did not enter a valid email, please try again.');
 }
 if($errors || !$_POST){

    if($errors){
        foreach($errors as $error){
            echo $error. "<br />";
        }
    }
}
if(!$errors && $success){
    mail($to, $subject, $message, $headers) && print($_POST['message']);

  } ?>
4

1 回答 1

0

首先将 else if 部分放在这样的括号中 ($_POST['message'] = filter_var($_POST['message'],FILTER_SANITIZE_STRING))

于 2013-01-26T18:24:25.243 回答