Filter and display portfolio items with multiple chekboxes

I am currently working on my protfolio for school. And i want to be able to filter and display my portfolio items by clicking 1 or more of the subjects, wich are chechboxes.

(for example like they did on this site: http://ift.tt/1N0sSSv )

I want to do this whitout clicking a submit button. But when a chekbox is checked or unchecked it wil "submit".

I had everything working until i wanted to auto-submit by chekking.

My code is like this:

            <form id="form" method="post" action="">
                <p><input class="chekbox" onchange="$('#form').submit();" type="checkbox" name="vak[]" value="SCO">SCO</p>
                <p><input class="chekbox" onchange="$('#form').submit();" type="checkbox" name="vak[]" value="UXU">UXU</p>
                <p><input class="chekbox" onchange="$('#form').submit();" type="checkbox" name="vak[]" value="DED">DED</p>
                <p><input class="chekbox" onchange="$('#form').submit();" type="checkbox" name="vak[]" value="PTT">PTT</p>
                <p><input class="chekbox" onchange="$('#form').submit();" type="checkbox" name="vak[]" value="PPM">PPM</p>

            </form>

So i removed the submit button because now it will auto submit. Yet if afther submitting and changing the displayed content the chekbox is unchecked. so if i check another checkbox it will directly submit and change display of newly checked box. So i cant select conect from multiple checkboxes.

Already tried to check box if vak isset. But then he checkes all boxes, because they are all same name. And i cant change name becuase then the wilde loop of displaying items wont work.

Here is the php

code

                 if(isset($_POST["vak"])){

                    if(!empty($_POST["vak"])){

                        foreach ($_POST["vak"] as $vak){

                            $select = "select * from portfolio where vak like '%$vak%'";
                            $run = mysqli_query($con, $select);

                            while ($row = mysqli_fetch_array($run)){

                                $id = $row['ID'];
                                $titel = $row['titel'];
                                $datum = $row['datum'];
                                $auteur = $row['auteur'];
                                $image = $row['image'];
                                $image2 = $row['image2'];
                                $image3 = $row['image3'];
                                $image4 = $row['image4'];
                                $keywords = $row['keywords'];
                                $inhoud = substr($row['beschrijving'],0,100);

I hope its claer what i want and whats not working and someone here can help me.

Thnx in advanced.