Question

So I have the submit button that looks like this:

<a href="#" id="submitbutton" 
onClick="document.getElementById('UploaderSWF').submit();">
<img src="../images/user/create-product.png" border="0" /></a>

When I double click it double submits obviously, and the problem is that I'm saving the information in the database so I'll have dublicate information there, and I dont want that. This uploader uses flash and javscript and here is a little piece of code that is relevant to the submit thing (if it helps)

$.fn.agileUploaderSubmit = function() {
    if($.browser.msie && $.browser.version == '6.0') {
        window.document.agileUploaderSWF.submit();
    } else {
        document.getElementById('agileUploaderSWF').submit();
        return false;
    }
}

Thank you guys. I appreciate your help. This is really something I was unable to do myself because I have such a little experience with js and I dont really know how to do stuff. THANKS.

Was it helpful?

Solution

Try this snipped:

$('#your_submit_id').click(function(){
    $(this).attr('disabled');
});

edit 1

Oh, in your case it is a link and no submit button ...

var submitted = false;

$.fn.agileUploaderSubmit = function() {
    if ( false == submitted )
    {
        submitted = true;

        if($.browser.msie && $.browser.version == '6.0') {
            window.document.agileUploaderSWF.submit();
        } else {
            document.getElementById('agileUploaderSWF').submit();
        }
    }

    return false;
}

edit 2

To simplify this, try this:

<!doctype html>

<html dir="ltr" lang="en">

<head>

<meta charset="utf-8" />

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"></script>

<script type="text/javascript">
<!--//--><![CDATA[//><!--
    $(document).ready(function()
    {
        $('#yourSubmitId').click(function()
        {
            $(this).attr('disabled',true);

            /* your submit stuff here */

            return false;
        });
    });
//--><!]]>
</script>

</head>
<body>

<form id="yourFormId" name="yourFormId" method="post" action="#">
    <input type="image" id="yourSubmitId" name="yourSubmitId" src="yourImage.png" alt="Submit" />
</form>​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​

</body>
</html>

Use form elements, like <input type="image" />, to submit a form not a normal link.

This works fine!

Take a look at jQuery.post() to submit your form.

Good luck.

edit 3

This works well for me too:

<!doctype html>

<html dir="ltr" lang="en">

<head>

<meta charset="utf-8" />

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"></script>

<script type="text/javascript">
<!--//--><![CDATA[//><!--
    $(document).ready(function()
    {
        var agileUploaderSWFsubmitted = false;

        $('#submitbutton').click(function()
        {
            if ( false == agileUploaderSWFsubmitted )
            {
                agileUploaderSWFsubmitted = true;

                //console.log( 'click event triggered' );

                if ( $.browser.msie && $.browser.version == '6.0' )
                {
                    window.document.agileUploaderSWF.submit();
                }
                else
                {
                    document.getElementById( 'agileUploaderSWF' ).submit();
                }
            }

            return false;
        });
    });
//--><!]]>
</script>

</head>
<body>

<form id="agileUploaderSWF" name="agileUploaderSWF" method="post" action="http://your.action/script.php">
    <input type="text" id="agileUploaderSWF_text" name="agileUploaderSWF_text" />
</form>​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​

<a href="#" id="submitbutton"><img src="../images/user/create-product.png" border="0" /></a>​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​

</body>
</html>

Hopefully this helps.

OTHER TIPS

Add the following to the onclick:

onclick="document.getElementById('submitbutton').disabled = true;document.getElementById('UploaderSWF').submit();"

That said, you will have to handle this double submit prevention on the server side also.

According to http://api.jquery.com/prop/ you can be add and remove the disabled property using the .prop() function, not the .attr() and .removeAttr() functions.

To add the property:

.prop("disabled", true);

To remove the property:

.prop("disabled", false);

Hope this will help you.

as proposed here (with examples):

if you want to make sure that the registered event is fired only once, you should use jQuery's [one][1] :

.one( events [, data ], handler ) Returns: jQuery

Description: Attach a handler to an event for the elements. The handler is executed at most once per element per event type.

This will disable all submit buttons and links with the class submit or the attribute data-submit in a form if one is clicked:

$(document).ready(function() {
    $('form').submit(function() {
        $(this).find('input[type="submit"]').attr('disabled', true);
        $(this).find('a[data-submit], a.submit').click(function() {
            return false;
        });
    }
});

This automatically applies to every form, by the way. If you just want a certain one, use the id instead of form. You most probably already knew that, though.

Licensed under: CC-BY-SA with attribution
Not affiliated with StackOverflow
scroll top