如何从php文件中获取消息并使用ajax发送到另一个php文件？

Question

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

        alert(msg); 

       } 
     }); 

我已经写了一个AJAX文件的一些参数发送到另一个文件insert.php发送这些参数作为邮件。如果邮件发送成功，它应该发送一条消息回主文件在那里显示。我已经写了如下图所示的insert.php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; 
} 
?> 

我想在$味精值发送回从那里控制来有显示它的PHP文件。如何获得成功函数内的消息？我试图提醒它，但它不起作用。

Answer 1

$.ajax({ 
    type: "POST", 
    url: "insert.php", 
    data: dataString, 
    cache: false, 
    success: function(result){ 
     alert(result);// the resiult is your message. 
    } 
}); 

无论你从insert.php回声结果是你的消息。请试试这个，你会得到结果。结果是该函数的参数。 ajax页面的输出只会在这里捕捉并仅处理这个结果。