如何从php文件获取消息并使用ajax发送到另一个php文件

$.ajax({
                type: "POST",
                url: "insert.php",
                data: dataString,
                cache: false,
                success: function(result){
                    alert(msg);
                }
        });

我已经写了一个ajax文件发送一些参数到另一个文件insert.php发送这些参数作为邮件。如果邮件发送成功，它应该向主文件发送一条消息以在主文件中显示它。我已经为insert.php编写了如下所示的代码:

<?php
 $date=$_POST['date'];
 $newDate = date("d-m-Y", strtotime($date ));
 $time=$_POST['time'];
 $name=$_POST['name'];
 $number=$_POST['number'];
 $mail=$_POST['mail'];
 $pname=$_POST['pname'];
 $ptype=$_POST['ptype'];
 $gender=$_POST['gender'];
 $comments=$_POST['comments'];
 $to  = 'df@gmail.com'; 
 $subject = 'Enquiry !!!!';

$message = ' A Customer has made a enquiry about a package  " '.$pname.'  "  whose Name is  "  '.$name.'  "  with Email-id   '.$mail.' and who    is  a  '.$gender.' and Contact Number is   '.$number.' and has send a message " '.$comments.' " on Date  '.$newDate .'  and Time  '.$time;
$headers  = 'MIME-Version: 1.0' . "'r'n";
$headers .= 'Content-type: text/html; charset=iso-8859-1' . "'r'n"; 
$retval = mail($to, $subject, $message, $headers);

if($retval == true)
{
  $msg = "YOUR ENQUIRY HAS BEEN SEND!!!!";
  echo $msg;
}else {
$msg = "YOUR ENQUIRY HAS NOT BEEN SEND!!!! PLEASE TRY AGAIN !!!";
echo $msg;
}
 ?>

我想发送$msg值回php文件从那里的控制来显示它。如何在成功函数中获得该信息?