0
$.ajax({
type: "POST",
url: "insert.php",
data: dataString,
cache: false,
success: function(result){
alert(msg);
}
});
我已經寫了一個AJAX文件的一些參數發送到另一個文件insert.php發送這些參數作爲郵件。如果郵件發送成功,它應該發送一條消息回主文件在那裏顯示。我已經寫了如下圖所示的insert.php代碼:如何從php文件中獲取消息並使用ajax發送到另一個php文件?
<?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 = '[email protected]';
$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 $msg;
}
?>
我想在$味精值發送回從那裏控制來有顯示它的PHP文件。如何獲得成功函數內的消息?我試圖提醒它,但它不起作用。
嘗試'alert(result)'。 –
不..不工作! – lena
我想提醒那裏的特定消息... – lena