PHP获取cookie的刷新值

onClick="javascript:document.cookie='title=< ?php echo $title; ? >';"   

<?php $tit=$_COOKIE['title']; ?>

正如@datasage指出onMouseDown在onClick之前被调用，所以你的addHit javascript在设置cookie值之前运行。详情请看此页

编辑:我明白你现在想做什么。你需要创建另一个php文件(在我的例子中是datagetter.php)来处理目录扫描并以一种有用的格式返回数据，我相信你想做的一切都不能在一个文件中同时发生，而不重新加载页面或在php代码的开头添加ajax处理程序。

使用php回显您的链接

<a href="ftp://<?php echo $source . '/' . $source2 . '/' . $genre . '/' . $title ?>" id="PopUpItUp" class="detail" data-title="<?php echo $title; ?>" data-id="<?php echo $id; ?>"><?php echo $title ?> (Episodes)</a>

使用jQuery为每个链接附加事件处理程序

<script>
$(window).load(function () {
    $('.detail').each().mousedown(function () {
        //set title cookie  access this in php with $_COOKIE['title']
        $.cookie('title', $(this).attr('data-title'));
    });
    $('.detail').each().click(function (e) {
        //increment hitcount
        var titleid = $(this).attr('data-id')
        addHit('./count.php?id=' + titleid , titleid);
        //get data to display via ajax call you may want this IN addHit somewhere
        //there will be a delay in getting the data you want so display something to show progress while we wait
        $("#popup").html("<div><img src='progresshappens.gif' width='25px' height='25px' />");
        $.ajax({
            url: 'http://yourdomain.com/datagetter.php?id=' + titleid,
            success: function(data) {
                //Do something with data, probably add to popup window? here is a starting point
                //Check into using javascript 'for each' function
                $('#popup').html('<div>' + data + '</div>');
            },
            error: function(request, error, errormessage) {
                //show error, something bad happened
                $("#messages").html(error + ''n' + errormessage);
            }
        });
        //don't follow link
        e.preventdefault();
        return false;
    });
</script>