Привет, есть сайт promojet.su есть всплывающая форма, работает на ajax,
после того как удачано отправили форму, выскакивает сообщение "спасибо" и если закрыть сообщение и опять вызвать форму не обновляя страницу, то формы не будет, а только сообщение "спасибо" .
Так вот вопрос, как вызвать по новой форму, не обновляя страницу.
код js:
$(document).ready(function() {
//if submit button is clicked
$('#form').submit(function () {
//Get the data from all the fields
var name = $('input[name=name]');
var email = $('input[name=email]');
var tel = $('input[name=tel]');
//Simple validation to make sure user entered something
//If error found, add hightlight class to the text field
if (name.val()=='') {
name.addClass('hightlight');
return false;
} else name.removeClass('hightlight');
if (email.val()=='') {
email.addClass('hightlight');
return false;
} else email.removeClass('hightlight');
if (tel.val()=='') {
tel.addClass('hightlight');
return false;
} else tel.removeClass('hightlight');
//organize the data properly
var data = 'name=' + name.val() + '&email=' + email.val() + '&tel=' +
tel.val();
//disabled all the text fields
//show the loading sign
$('.loading').show();
//start the ajax
$.ajax({
//this is the php file that processes the data and send mail
url: "process.php",
//GET method is used
type: "GET",
//pass the data
data: data,
//Do not cache the page
cache: false,
//success
success: function (html) {
//if process.php returned 1/true (send mail success)
if (html==1) {
//hide the form
//show the success message
$('.blk-data .form1').fadeOut('slow');
$('.loading').fadeOut('slow');
$('.done').fadeIn('slow');
//if process.php returned 0/false (send mail failed)
} else alert('Sorry, unexpected error. Please try again later.');
}
});
//cancel the submit button default behaviours
return false;
});
});
код php
//Retrieve form data.
//GET - user submitted data using AJAX
//POST - in case user does not support javascript, we'll use POST instead
$name = ($_GET['name']) ? $_GET['name'] : $_POST['name'];
$email = ($_GET['email']) ?$_GET['email'] : $_POST['email'];
$tel = ($_GET['tel']) ?$_GET['tel'] : $_POST['tel'];
//flag to indicate which method it uses. If POST set it to 1
if ($_POST) $post=1;
//Simple server side validation for POST data, of course, you should validate the email
if (!$name) $errors[count($errors)] = 'Please enter your name.';
if (!$email) $errors[count($errors)] = 'Please enter your email.';
if (!$tel) $errors[count($errors)] = 'Please enter your tel.';
//if the errors array is empty, send the mail
if (!$errors) {
//recipient
$to = 'Your Name <hello@domain.ru>';
//sender
$from = $name . ' <' . $email . '>';
//subject and the html message
$subject = 'Сообщение с сайта PromoJet ';
$message = '
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head></head>
<body>
<table>
<tr><td><strong>Имя:</strong></td><td>' . $name . '</td></tr>
<tr><td><strong>Email:</strong></td><td>' . $email . '</td></tr>
<tr><td><strong>Телефон:</strong></td><td>' . $tel . '</td></tr>
</table>
</body>
</html>';
//send the mail
$result = sendmail($to, $subject, $message, $from);
//if POST was used, display the message straight away
if ($_POST) {
if ($result) echo 'Thank you! We have received your message.';
else echo 'Sorry, unexpected error. Please try again later';
//else if GET was used, return the boolean value so that
//ajax script can react accordingly
//1 means success, 0 means failed
} else {
echo $result;
}
//if the errors array has values
} else {
//display the errors message
for ($i=0; $i<count($errors); $i++) echo $errors[$i] . '<br/>';
echo '<a href="/">Back</a>';
exit;
}
//Simple mail function with HTML header
function sendmail($to, $subject, $message, $from) {
$headers = "MIME-Version: 1.0" . "\r\n";
$headers .= "Content-type:text/html;charset=iso-8859-1" . "\r\n";
$headers .= 'From: ' . $from . "\r\n";
$result = mail($to,$subject,$message,$headers);
if ($result) return 1;
else return 0;
}
