【发布时间】:2014-03-04 10:16:26
【问题描述】:
我想在我的网站上添加一个弹出式 Fb 框,当访问者向下滚动时会出现弹出式窗口。我希望当访问者单击关闭按钮时弹出框不会出现 15 天。但是当我单击它的关闭时,它会在我向上或向下滚动时显示。请问有人可以帮我吗? 我的代码是:
<!-- FB Popup Likebox Start -->
<div class="fb-pop-like-box" id="fb-popup-like-box">
<p style="text-align: center;">Hey, Like Us on Facebook!</p>
<div class="fb-like-box" data-href="https://www.facebook.com/ifo4all" data-width="300" data-height="200" data-colorscheme="light" data-show-faces="true" data-header="false" data-stream="false" data-show-border="true"></div>
<div class="no-show"><a href="#">Close: I already Like if04all!</a></div>
</div>
<script type="text/javascript">
$(function () {
$("#fb-popup-like-box").hide();
$(window).scroll(function () {
if ($(this).scrollTop() > 2100) {
$('#fb-popup-like-box').slideDown('slow');
} else {
$('#fb-popup-like-box').slideUp('slow');
}
});
});
jQuery(document).ready(function(){
if(readFBCookie('squenched') != null) {
jQuery('#fb-popup-like-box').css("display", "none");
}
jQuery('.no-show a').live("click",function(event){
event.preventDefault();
jQuery('#fb-popup-like-box').css("display", "none");
var days=15;
var date = new Date();
date.setTime(date.getTime()+(days*24*60*60*1000));
var expires = "; expires="+date.toGMTString();
document.cookie = "squenched=1"+expires+"; path=/";
return false;
});
function readFBCookie(name) {
var nameEQ = name + "=";
var ca = document.cookie.split(';');
for(var i=0;i < ca.length;i++) {
var c = ca[i];
while (c.charAt(0)==' ') c = c.substring(1,c.length);
if (c.indexOf(nameEQ) == 0) return c.substring(nameEQ.length,c.length);
}
return null;
}
});
</script>
<!-- FB Popup Likebox Start End -->
【问题讨论】:
标签: javascript jquery facebook cookies popup