AJAX jQuery…变量未传递给PHP

大家好,
尝试做一些非常简单的事情：将2个文本变量传递给PHP脚本,并将其插入到MySQL数据库中.但是由于某种原因,我无法传递变量(因此,我在数据库中只得到空记录).

function ajaxCall(){

        $.ajax({
            type: "GET",
            url: "http://www.*.be/bubblingAjax.PHP",
            cache: false,
            data: "colour="+colour+"&size="+size,
        dataType: "html",
            success: onSuccess
        });
        return false;
    };

和PHP：

<?PHP
    try
    {
        $connection = MysqL_connect("#");
        MysqL_select_db("#");

        $colour = MysqL_real_escape_string($_GET['colour']);
        $size = MysqL_real_escape_string($_GET['size']);

        MysqL_query("INSERT INTO bubble (colour, size) VALUES ('$colour', '$size')");
        MysqL_close($connection);
        echo "SUCCESS";
        echo $colour;
        echo $size;
    }
    catch(Exception $e)
    {
        echo $e->getMessage();
    }
?>

任何人愿意快速浏览一下并指出我的-可能是显而易见的-错误吗？一天来一直让我发疯！

谢谢！

解决方法:

这必须工作：

<script type="text/javascript">
$(document).ready(function() {
   //you can wrap the code into an event, e.g click()
   var colour=...
   var size=...
   $.post("http://www.website.com/bubblingajax.PHP", { colour: colour, size: size },
   function(data) {
     alert("Respond: " + data);
   }); 
});

</script>

和PHP(只有更改才能发布)

<?PHP
    try
    {
        $connection = MysqL_connect("#");
        MysqL_select_db("#");

        $colour = MysqL_real_escape_string($_POST['colour']);
        $size = MysqL_real_escape_string($_POST['size']);

        MysqL_query("INSERT INTO bubble (colour, size) VALUES ('$colour', '$size')");
        MysqL_close($connection);
        echo "SUCCESS";
        echo $colour;
        echo $size;
    }
    catch(Exception $e)
    {
        echo $e->getMessage();
    }
?>

同样为了调试,我建议在检查工具中使用firebug或chrome的构建.