Php 邮件不适用于多个收件人
Php mail not working with more than one recipient
为什么 this piece of php code 不适用于多个收件人?
只有当 $to
只有一个地址时才有效,例如:
$to = 'aa@bb.com';
编辑
如果电子邮件地址在同一个域中,它会起作用。例如,如果网站是 www.example.com
,xxx@example.com
之类的电子邮件会起作用,但 yyy@other.com
不会.
解决方案
PHPMailer。它提供了一种简单的方法来配置 SMTP
.
这里是初始代码
<?php
//define the receiver of the email
$to = 'aa@bb.com, cc@dd.com, ee@ff.com';
// array with filenames to be sent as attachment
$files = array("a.zip","b.c","a.html");
// email fields: to, from, subject, and so on
$from = "mail@mail.com";
$subject ="My subject";
$message = "My message";
// boundary
$semi_rand = md5(time());
$mime_boundary = "==Multipart_Boundary_x{$semi_rand}x";
// headers for attachment
$headers = "From: $from";
$headers .= "\nMIME-Version: 1.0\n" . "Content-Type: multipart/mixed;\n" . " boundary=\"{$mime_boundary}\"";
// multipart boundary
$message = "This is a multi-part message in MIME format.\n\n" . "--{$mime_boundary}\n" . "Content-Type: text/plain; charset=\"iso-8859-1\"\n" . "Content-Transfer-Encoding: 7bit\n\n" . $message . "\n\n";
$message .= "--{$mime_boundary}\n";
// preparing attachments
for($x=0;$x<count($files);$x++){
$file = fopen($files[$x],"rb");
$data = fread($file,filesize($files[$x]));
fclose($file);
$data = chunk_split(base64_encode($data));
$message .= "Content-Type: {\"application/octet-stream\"};\n" . " name=\"$files[$x]\"\n" .
"Content-Disposition: attachment;\n" . " filename=\"$files[$x]\"\n" .
"Content-Transfer-Encoding: base64\n\n" . $data . "\n\n";
$message .= "--{$mime_boundary}\n";
}
// send
$ok = @mail($to, $subject, $message, $headers);
if ($ok) {
echo "<p>mail sent to $to!</p>";
} else {
echo "<p>mail could not be sent!</p>";
}
?>
这是最终代码
// PHPMailer
$mail = new PHPMailer;
// setting up PHPMailer
$mail->isSMTP();
$mail->Host = 'host.com';
$mail->SMTPAuth = false;
$mail->Port = xx;
$mail->setFrom($_POST['email'], $_POST['name']);
$mail->Subject = $_POST['subject'];
$mail->msgHTML($_POST['message']);
foreach($contacts as $contact)
$mail->addAddress($contact['email']);
// If the user has attached some files
foreach ($_FILES as $file)
$mail->addAttachment($file['tmp_name'], basename($file['name']));
$response = array("status" => $mail->send() ? "sent" : "error");
echo json_encode($response);
您需要使用正确的 RFC 2822 格式。
不要使用@,因为您不知道错误是什么。如果您将邮件格式化为 "user@example.com, anotheruser@example.com" 格式,这是正确的,您需要在别处搜索问题。
您还可以在 http://php.net/manual/en/function.mail.php#example-3493 页面上查看示例 #4。
虽然这不是对您的问题的直接回应,但您可以通过使用预先存在的电子邮件发送库来避免麻烦,例如:
phpmailer
https://github.com/PHPMailer/PHPMailer
为什么 this piece of php code 不适用于多个收件人?
只有当 $to
只有一个地址时才有效,例如:
$to = 'aa@bb.com';
编辑
如果电子邮件地址在同一个域中,它会起作用。例如,如果网站是 www.example.com
,xxx@example.com
之类的电子邮件会起作用,但 yyy@other.com
不会.
解决方案
PHPMailer。它提供了一种简单的方法来配置 SMTP
.
这里是初始代码
<?php
//define the receiver of the email
$to = 'aa@bb.com, cc@dd.com, ee@ff.com';
// array with filenames to be sent as attachment
$files = array("a.zip","b.c","a.html");
// email fields: to, from, subject, and so on
$from = "mail@mail.com";
$subject ="My subject";
$message = "My message";
// boundary
$semi_rand = md5(time());
$mime_boundary = "==Multipart_Boundary_x{$semi_rand}x";
// headers for attachment
$headers = "From: $from";
$headers .= "\nMIME-Version: 1.0\n" . "Content-Type: multipart/mixed;\n" . " boundary=\"{$mime_boundary}\"";
// multipart boundary
$message = "This is a multi-part message in MIME format.\n\n" . "--{$mime_boundary}\n" . "Content-Type: text/plain; charset=\"iso-8859-1\"\n" . "Content-Transfer-Encoding: 7bit\n\n" . $message . "\n\n";
$message .= "--{$mime_boundary}\n";
// preparing attachments
for($x=0;$x<count($files);$x++){
$file = fopen($files[$x],"rb");
$data = fread($file,filesize($files[$x]));
fclose($file);
$data = chunk_split(base64_encode($data));
$message .= "Content-Type: {\"application/octet-stream\"};\n" . " name=\"$files[$x]\"\n" .
"Content-Disposition: attachment;\n" . " filename=\"$files[$x]\"\n" .
"Content-Transfer-Encoding: base64\n\n" . $data . "\n\n";
$message .= "--{$mime_boundary}\n";
}
// send
$ok = @mail($to, $subject, $message, $headers);
if ($ok) {
echo "<p>mail sent to $to!</p>";
} else {
echo "<p>mail could not be sent!</p>";
}
?>
这是最终代码
// PHPMailer
$mail = new PHPMailer;
// setting up PHPMailer
$mail->isSMTP();
$mail->Host = 'host.com';
$mail->SMTPAuth = false;
$mail->Port = xx;
$mail->setFrom($_POST['email'], $_POST['name']);
$mail->Subject = $_POST['subject'];
$mail->msgHTML($_POST['message']);
foreach($contacts as $contact)
$mail->addAddress($contact['email']);
// If the user has attached some files
foreach ($_FILES as $file)
$mail->addAttachment($file['tmp_name'], basename($file['name']));
$response = array("status" => $mail->send() ? "sent" : "error");
echo json_encode($response);
您需要使用正确的 RFC 2822 格式。
不要使用@,因为您不知道错误是什么。如果您将邮件格式化为 "user@example.com, anotheruser@example.com" 格式,这是正确的,您需要在别处搜索问题。
您还可以在 http://php.net/manual/en/function.mail.php#example-3493 页面上查看示例 #4。
虽然这不是对您的问题的直接回应,但您可以通过使用预先存在的电子邮件发送库来避免麻烦,例如:
phpmailer https://github.com/PHPMailer/PHPMailer