PHP梅勒您必须至少提供一个电子邮件地址

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; 
    } 
} 

我传递其持有这种格式的电子邮件地址的字符串：

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

当我路过这个

$recipient = array ($to); 

我有错误无效的地址：但是当我直接传递字符串输出如下：

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

它工作正常。如何将我的$传递给这个$收件人数组。

Answer 1

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

不会奇迹般地创建一个包含两个元素的数组。什么，你会得到的是数组中的一个元素，如

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

但是，这样做

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

会给你两个元素的数组：

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