Reputation: 5647
I am using .submit()
to submit my form over ajax periodically, but I want the user to see that the form is being saved with a spinning wheel and then 'Saved!' upon success. Is there a success
trigger for .submit()
in jQuery?
Thanks!
Upvotes: 12
Views: 50819
Reputation: 2277
Try this:
jQuery:
$(document).ready(function() {
$('#form').submit(function() {
var status = '<img class="loading" src="loading_detail.gif" alt="Loading..." />';
$("#ajax").after(status);
$.ajax({
type: 'POST',
url: $(this).attr('action'),
data: $(this).serialize(),
dataType: 'json',
success: function(json) {
if(json.type == 'success') {
$('#msg').css("color","green").html(json.message);
} else if(json.type == 'warning'){
$('#msg').css("color","yellow").html(json.message);
} else if(json.type == 'error'){
$('#msg').css("color","red").html(json.message);
}
$('.loading').remove();
}
})
return false;
});
});
Html:
<div id="msg"></div>
<form id="form" method="post" action="action.php" >
<input type="text" name="email" />
<input type="submit" name="submit" value="submit" /><span id="ajax"></span>
</form>
action.php :
<?php
if(isset($_POST['email'])){
$val = $_POST['email'];
switch ($val) {
case '[email protected]':
$return = array('type'=>'success', 'message'=>'This is success message!'); break;
case 'email':
$return = array('type'=>'warning', 'message'=>'This is warning message!'); break;
default:
$return = array('type'=>'error', 'message'=>'This is error message!');
}
echo json_encode($return);
}
?>
Note:
If you are submitting the form programmatically you need to call $('#form').submit()
once again but this time without arguments so that you trigger the submit event.
Upvotes: 19
Reputation: 19
To show a spinning wheel while submitting a page, try to get a animated GIF (there are much free loading circles). This GIF implement in your page with:
class="myloadingcircle" style="display:none"
and then use jQuery:
$("form").submit(function(e) { $(".myloadingcircle").show(); });
Upvotes: 1
Reputation: 298176
You can use a manual $.post()
request instead of a .submit()
. The $.post()
has a success callback.
You'll have to fill in the details of the $.post()
(the target URL) and .serialize()
your form's elements.
Upvotes: 7