jQuery ajax导致页面刷新

时间:2011-11-09 04:15:10

标签: jquery ajax

当用户滚动到页面底部时,它会导致我的数据库中的某些项目Lazy loading

以下是标题中的js:

<script type="text/javascript">
var counter = 0;
var categoryid = 93;
        $(window).scroll(function(){
            if ($(window).scrollTop() >= $(document).height() - $(window).height() - 10) {
                $('div#loadmoreajaxloader').show();
                counter = counter + 20;
                $.ajax({
                    url: "loadmore.php?counter=" + counter + "&categoryid=" + categoryid,
                    success: function(html){
                        if(html){
                            $("#postswrapper").append(html);
                            $('div#loadmoreajaxloader').hide();
                        }else{
                            $('div#loadmoreajaxloader').html('<center>No more coupons to show.</center>');
                        }
                    }
                });
            }
        });
</script>

问题是,当我到达页面底部时,它会尝试刷新自己。但是我只是得到一个白色的屏幕。我不明白发生了什么???

当我转到loadmore.php?counter=20&categoryid=93页面时,它会正常加载并输出我想要的html。

好的,我只是在输出一些非常简单的html时才能使用懒惰的工作。看起来问题在于它试图输出的内容。这就是我想要输出的内容,它有一些javascript,它看起来像是通过ajax导致一些问题:

    <?php

    echo "<div class='category rounded_box'>
    <div class='category-thumbnail'>
        <a href='http://www.mydomain.com/site/kensie.com' title='View more Kensie Coupon Codes'>
            <img src='http://static.mydomain.com/images/logo-small.jpg' width='88' height='31'
                border='0' alt='Kensie - Logo' class='logo' /></a></div>
    <div class='coupon_details_homepage'>
        <div class='category_title'>
            <div class='coupon_title_homepage'>
                40% off when you share with your friends.
            </div>
        </div>
        <div class='calltoaction'>
            <a class='box' onclick='popup(http://www.mydomain.com/go/23147)' rel='nofollow' href='#coupon-23147'>
                Click Here to Activate Coupon </a>
        </div>
        <div style='clear: both; margin-top: 5px;'>
            More <a href='http://www.mydomain.com/site/kensie.com' title='View more Kensie Coupon Codes'>
                Kensie Coupon Codes</a>
        </div>
    </div>
</div>
<script type='text/javascript'>showtext('By clicking this link you have activated the coupon. View the newly opened Kensie window to view this deal!','coupon-23147')</script>";
        ?>

我已将其缩小到导致问题的原因。在html的末尾,我需要输出一个与输出链接交互的javascript函数。有办法解决这个问题吗?

<script type='text/javascript'>showtext('By clicking this link you have activated the coupon. View the newly opened Kensie window to view this deal!','coupon-23147')</script>

0 个答案:

没有答案