包括密件抄送和收件人的联系方式

Question

我想发送密件抄送电子邮件以及联系人表单中的收件人。如何在我的代码中包含密件抄送？我在下面的代码中没有看到错误，但它不会向您发送电子邮件至密件抄送或收件人。如果我删除密件抄送，它会将您发送给收件人。

 $recipient = "xjkfak223@gmail.com";   

// Set the email subject.
$subject = "New contact from $name";

// Build the email content.
$email_content = "Name:  $name\n";
$email_content .= "Email:  $email\n\n";
$email_content .= "Phone:  $phone\n\n";
$email_content .= "Message:  $message\n";
$email_content .= "$subscribe: Yes\n";

// Build the email headers.
$email_headers = "From: $name <$email>";
$email_headers = "BCC: xy123@gmail.com";

// Send the email.
if (mail($recipient, $subject, $email_content, $email_headers)) {
    // Set a 200 (okay) response code.
    http_response_code(200);
    echo "Thank You for Contacting Us.";
} 

Answer 1

您正在覆盖变量。
如果您添加.=，它将附加该字符串。我想你也需要一个 comma 换行符。

$email_headers = "From: $name <$email>\n";
$email_headers .= "BCC: xy123@gmail.com";