【发布时间】:2015-07-05 02:14:56
【问题描述】:
//send email
$to = $_POST['email'];
$subject = "Welcome!";
$body = "Contains sensitive information that activates users so I've removed it.";
$additionalheaders = "From: <".SITEEMAIL.">\r\n";
$additionalheaders .= "Reply-To: ".SITEEMAIL."";
mail($to, $subject, $body, $additionalheaders);
基本上,上面的代码成功发送了一封电子邮件,并且在收件箱中显示为 noreply@mydomain,这很好,但我注意到其他网站显示的是实际名称。像来自 Facebook 的消息说 Facebook 不是 noreply@facebook.com。是否有我缺少的标题来完成此操作?
编辑显示答案:
//send email
$to = $_POST['email'];
$subject = "Welcome!";
$body = "Contains sensitive information that activates users so I've removed it.";
$additionalheaders = "From: Name <".SITEEMAIL.">\r\n";
$additionalheaders .= "Reply-To: ".SITEEMAIL."";
mail($to, $subject, $body, $additionalheaders);
【问题讨论】:
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$additionalheaders = "来自:\r\n";在这里想吗?