PHP 电子邮件表单发件人姓名而不是电子邮件？

Question

现在，我已经设置了一个 PHP 电子邮件表单，一切正常。 但是，在测试它时，我只将发件人电子邮件地址作为名称。 我想要的是发件人的姓名，例如：

约翰·多伊

主题 主题 主题

代替：

j.doe@website.com

主题 主题 主题

下面是代码...

有人能帮助我吗？ 谢谢。

PHP：

<?php
if(isset($_POST['email'])) {

    // EDIT THE 2 LINES BELOW AS REQUIRED
    $email_to = "ricky@rickytsang.ca";
    $email_subject = $_REQUEST['subject'];


    function died($error) {
        // your error code can go here
        echo "We are very sorry, but there were error(s) found with the form you submitted. ";
        echo "These errors appear below.<br /><br />";
        echo $error."<br /><br />";
        echo "Please go back and fix these errors.<br /><br />";
        die();
    }

    // validation expected data exists
    if(!isset($_POST['full_name']) ||
        !isset($_POST['email']) ||
        !isset($_POST['subject']) ||
        !isset($_POST['message'])) {
        died('We are sorry, but there appears to be a problem with the form you submitted.');      
    }

    $full_name = $_POST['full_name']; // required
    $email_from = $_POST['email']; // required
    $subject = $_POST['subject'];
    $message= $_POST['message']; // required

    $error_message = "";
    $email_exp = '/^[A-Za-z0-9._%-]+@[A-Za-z0-9.-]+\.[A-Za-z]{2,4}$/';
  if(!preg_match($email_exp,$email_from)) {
    $error_message .= 'The e-mail you entered does not appear to be valid.<br />';
  }
    $string_exp = "/^[A-Za-z .'-]+$/";
  if(!preg_match($string_exp,$full_name)) {
    $error_message .= 'The name you entered does not appear to be valid.<br />';
  }
  if(strlen($message) < 2) {
    $error_message .= 'The message you entered doee not appear to be valid.<br />';
  }
  if(strlen($error_message) > 0) {
    died($error_message);
  }
    $email_message = "Form details below.\n\n";

    function clean_string($string) {
      $bad = array("content-type","bcc:","to:","cc:","href");
      return str_replace($bad,"",$string);
    }

    $email_message .= "Full Name: ".clean_string($full_name)."\n";
    $email_message .= "E-mail: ".clean_string($email_from)."\n";
    $email_message .= "Subject: ".clean_string($subject)."\n";
    $email_message .= "Message: ".clean_string($message)."\n";


// create email headers
$headers = 'From: '.$email_from."\r\n".
'Reply-To: '.$email_from."\r\n" .
'X-Mailer: PHP/' . phpversion();
@mail($email_to, $email_subject, $email_message, $headers); 
?>

<!-- include your own success html here -->

Thank you for contacting us. We will be in touch with you very soon.

<?php
}
?>

Answer 1

在你说$email_message.=...的 4 行之后添加一行：

$email_from = $full_name.'<'.$email_from.'>';

Answer 2

那么你需要这种形式：

"John Doe <j.doe@website.com>"

Answer 3

通常你只需要像这样重新格式化发件人地址：

"Ricky, Dear" <dearricky@yahoo.com>

http://php.net/manual/en/function.mail.php

Answer 4

谢谢你！

因此，您的整个代码部分将如下所示：

// prepare email body text
$message = "";

$message .= "Thanks for registering .....";

$message .= "\n";

$message .= "More stuff here";

$headers = "From: John Smith <info@yourcoolsite.com>";


mail($recipientemail, "Hey thanks for joining!", $message, $headers);

然后 email 收件人将在收件箱中看到 email，如下所示：

来自：约翰史密斯

主题：嘿，谢谢你的加入！

MESSAGE: 感谢您的注册.....更多的东西在这里

Answer 5

在标题中，您必须像这样设置 From header：

<?php
...
$headers = 'From: ' . clean_string($full_name) . ' <' . clean_string($email_from) . ">\r\n";
..
?>

Answer 6

您这样设置电子邮件地址

$email_to = "Ricky <ricky@rickytsang.ca>";

http://php.net/manual/en/function.mail.php

Answer 7

更改这 4 行

 $full_name = $_POST['full_name'];
    $email_from = $_POST['email'];
    $subject = $_POST['subject'];
    $message= $_POST['message'];

至：

 $full_name = $_POST['full_name'];
    $email_from = $_POST['email'];
    $subject = $_POST['subject'];
    $message= '"' . $full_name . '" <' . $_POST['message'] . '>';

请确保您照原样复制粘贴。

Answer 8

Drupal 电子邮件表单发件人姓名而不是基于域的电子邮件？

例子：

function hook_mail_alter(&$message) {

if($message['id'] == "put_id" && $domain_id == 368){

$default_from = variable_get('site_mail', ini_get('sendmail_from'));

if($message['from'] == $default_from){

$message['from'] = '"'. variable_get('site_name', 'Drupal') .'" <'. 
$default_from .'>';

$message['headers']['From'] = $message['from'];

}

}

}