Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Show loading gif after clicking form submit using jQuery

I'm trying to simply show a loading gif once the form is submitted. My code is not working correctly for the loading gif to show. You'll see my image for the gif is set to visibility: hidden.

<script src="js/jquery-1.11.1.min.js"></script>

<div class="" style="width: 480px; margin: 0 auto; margin-top: 20px;" id="form_wrapper">
    <img src="img/loader.gif" id="gif" style="display: block; margin: 0 auto; width: 100px; visibility: hidden;">
    <div class="fade" id="form">
        <form action="process_login.php" method="POST" name="simplelogin" id="login_form"><br />
            <label for="username">&nbsp; Username:</label>
            <input type="username" name="username" id="username" size="30" required><br><br>
            <label for="password">&nbsp; Password:</label>
            <input type="password" name="password" id="password" size="30" required><br><br>
            <input class="load_button" type="submit" name="submit" id="submit" value="Submit" placeholder="Submit">
        </form>
    </div>
</div>
<div class="fifty"></div>

<script type="text/javascript">
    $('.load_button').submit(function() {
        $('#gif').show(); 
        return true;
    });
</script>
like image 531
ValleyDigital Avatar asked Nov 19 '14 20:11

ValleyDigital


3 Answers

The show() method only affects the display CSS setting. If you want to set the visibility you need to do it directly. Also, the .load_button element is a button and does not raise a submit event. You would need to change your selector to the form for that to work:

$('#login_form').submit(function() {
    $('#gif').css('visibility', 'visible');
});

Also note that return true; is redundant in your logic, so it can be removed.

like image 181
Rory McCrossan Avatar answered Nov 14 '22 23:11

Rory McCrossan


Button inputs don't have a submit event. Try attaching the event handler to the form instead:

<script type="text/javascript">
     $('#login_form').submit(function() {
       $('#gif').show(); 
       return true;
     });
 </script>
like image 8
Dave Avatar answered Nov 14 '22 23:11

Dave


Better and clean example using JS only

Reference: TheDeveloperBlog.com

Step 1 - Create your java script and place it in your HTML page.

<script type="text/javascript">
    function ShowLoading(e) {
        var div = document.createElement('div');
        var img = document.createElement('img');
        img.src = 'loading_bar.GIF';
        div.innerHTML = "Loading...<br />";
        div.style.cssText = 'position: fixed; top: 5%; left: 40%; z-index: 5000; width: 422px; text-align: center; background: #EDDBB0; border: 1px solid #000';
        div.appendChild(img);
        document.body.appendChild(div);
        return true;
        // These 2 lines cancel form submission, so only use if needed.
        //window.event.cancelBubble = true;
        //e.stopPropagation();
    }
</script>

in your form call the java script function on submit event.

<form runat="server"  onsubmit="ShowLoading()">
</form>

Soon after you submit the form, it will show you the loading image.

like image 3
BenW Avatar answered Nov 14 '22 21:11

BenW