Question

require_once('phpmailer/class.phpmailer.php');

function smtpmailer($to,$from,$subject,$body) { 
    define('GUSER', 'xxx'); // Gmail username
    define('GPWD', 'xxx'); // Gmail password
    printf("list:".$to);
    $recipient = array ($to);
    global $error;
    $mail = new PHPMailer();  // create a new object
    $mail->IsSMTP(); // enable SMTP
    $mail->SMTPDebug = 1;  // debugging: 1 = errors and messages, 2 = messages only
    $mail->SMTPAuth = true;  // authentication enabled
    $mail->SMTPSecure = 'ssl'; // secure transfer enabled REQUIRED for Gmail
    $mail->Host = 'smtp.gmail.com';
    $mail->Port = 465; 
    $mail->Username = GUSER;  
    $mail->Password = GPWD;           
    $mail->SetFrom($from, "Bank Negara");
    $mail->Subject = $subject;
    $mail->Body = $body;
    foreach ($recipient as $email){
    $mail->AddAddress($email);
    }
    if(!$mail->Send()) {
        $error = 'Mail error: '.$mail->ErrorInfo; 
        return false;
    } else {
        $error = 'Message sent!';
        return true;
    }
}

我传递的字符串以这种格式保存电子邮件地址：

'email1@yahoo.com','email2@gmail.com'

当我通过这个

时

$recipient = array ($to);

我遇到错误地址无效：但是当我像这样直接传递字符串输出时：

$recipient = array ('email1@yahoo.com','email2@gmail.com');

工作正常。应该如何将$ $传递给这个$ recipient数组。

Answer 1

$addresses = "'email1@yahoo.com','email2@gmail.com'";
$to = array($addresses);

不会神奇地创建一个包含两个元素的数组。你得到的是数组中的SINGLE元素，例如

$to = array(
    0 => "'email1@yahoo.com','email2@gmail.com'"
);

然而，做

$to = explode(',', $addresses);

会给你一个两元素数组：

$to = array (
    0 => "...yahoo",
    1 => "...gmail"
);

PHP邮件程序您必须至少提供一个电子邮件地址

1 个答案: