php database, jquery dynamic content load - php

Question

I would like to achieve such an effect, except that when you click on the picture (link) jquery script sends a GET to the same file from the fact that with another ID. I mean the dynamic reload the page without refreshing the addition of a nice effect in the attached link.
my code :
<script>
$('a.menu').click(function(){
$('.content').html('');
})
</script>
<div class="content" id="page_effect" style="padding:0px; display:none;">
<div class="separator" style="margin: -17px auto;"></div>
<span class="choose-product"> Wybierz Produkt</span>
<p>
<?php
$kat=$_GET['kat'];
$co_ile_strona=9;
//----------------
$dopisz="";
if (is_numeric($kat)) {
$dopisz=" WHERE kat_id='".$kat."'";
$wyk=mysql_query("SELECT * FROM kategorie WHERE kat='".$kat."'");
while($ww=mysql_fetch_array($wyk)) {
$dopisz.=" OR kat_id='".$ww['id']."'";
}
}
$sile=false;
$wyk=mysql_query("SELECT * FROM produkty ".$dopisz."");
if ($ile=mysql_num_rows($wyk)) {
if (!$sile) {
$nazwa = mysql_fetch_assoc(mysql_query("SELECT * FROM kategorie WHERE id='".$_GET['kat']."'"));
if(strlen($nazwa['nazwa']) > 0)
$nazwa = $nazwa['nazwa'];
?>
<div style="text-align: center; width: 80%;margin: 0 auto;margin-top: 39px;">
<a href="produkt.html"><div class="product-box">
<img src="images/picasso0.png" alt="Product"/>
<span class="product-title"><?=$nazwa?></span>
</div>
</a>
<?
$sile=true;
}
if (!$_GET['strona']) $strona=1; else $strona=$_GET['strona'];
$start=($strona*$co_ile_strona)-$co_ile_strona;
mysql_data_seek($wyk,$start);
$licz=0;
while(($ww=mysql_fetch_object($wyk)) && $licz<$co_ile_strona) { $licz++;
?>
<a href="<?=strtolower(seo($ww->nazwa))?>-<?=$ww->id?>p.html"><div class="product-box">
<img src="produkty/front/<?=$ww->front?>" alt="<?=$ww->nazwa?>"/>
<div class="name2"><span><?=$ww->nazwa?> </span></div>
</div>
</a>
<?
}
} else echo "<span style='color: #ff0000; font-size: 12pt; font-weight: bold;'>Przepraszamy, ale nie znaleziono produktów pasujących do tego zapytania</span>";
?>
</div>
<div class="menu-bottom" style="text-align:center;">
<span style="position: relative;top: 25px;display: inline-flex;margin-bottom: 20px;">Wybierz serię:
<ul>
<?php
$zapas=$_GET['kat'];
$wyk=mysql_query("SELECT * FROM kategorie WHERE kat='0' and wid='1' ORDER BY poz ASC");
while($ww=mysql_fetch_object($wyk)) {
?> <!--<?/*=$ww->nazwa?>-<?=$ww->id*/?>k.html*/-->
<li> <? if($_GET['kat']==$ww->id) echo "<span style='color: #000;'>".$ww->nazwa.""; else echo $ww->nazwa?></li>
<? } ?>
</ul>
</span>
</div>
<!-- end .content --></div>
link : Click here

Answer 1

As mentioned in the comment, you should send an AJAX request to the page that is responsible for handling the database tasks. You can use get() or post() function which is a shorthand of AJAX function as stated in jQuery documentation.
$.ajax({
url: url,
data: data,
success: success,
dataType: dataType
});
I've prepared a simple jsFiddle demonstrating how the task can be achieved (please note the event.preventDefault() call):
$(document).ready(function(){
$("button").click(function(e){
var URL = '';
$.get(URL,function(data){
console.log("Status: " + status);
e.preventDefault();
});
});
});
Hope that helps.