如何在访问者留在网站3分钟后显示弹出窗口,它可以是网站的任何页面。
我读过一篇文章来使用setTimeout(function(){},time); 我是JS的新手但每次添加Timeout都会破坏我的代码。
感谢您的帮助!!
我的代码在
下面 <div id="popup_a" class="popup_ABA">
<table width="100%" border="0" align="center" cellpadding="0" cellspacing="0">
<tr>
<td align="center"><div>THANK YOU FOR YOUR HELP</div></td>
</tr>
<tr>
<td><table width="100%" border="0" cellspacing="0" cellpadding="0">
<tr>
<td width="46%" align="right"><a href="#" class="nothanks">No</a></td>
<td width="8%" class="popUpFooter_ABA"> </td>
<td width="46%" align="left"><span class="popUpFooter_ABA"><a href="#" class="dosurvey" onClick="dosurvey('#', 'invitationwindow', 800, 550);">Yes</a></span></td>
</tr>
</table></td>
</tr>
</table>
</div>
<script type="text/javascript">
$(document).ready( function() {
$.cookie('cisit292014', '1', {expires: 15, path: '/'});
setInterval(function() {
if ($.cookie('cisit292014') == 11){
if ($.cookie('01092014') != '1') { // Check for cookie
$.cookie('01092014', '1', { expires: 30, path: '/' });
// TO Unload the Popupbox
function unloadPopupBox(){$('#popup_a').fadeOut("slow");}
// To Load the Popupbox
function loadPopupBox(){$('#popup_a').fadeIn("slow");}
// When site loaded, load the Popupbox First
loadPopupBox();
$('.nothanks').click( function() {unloadPopupBox();});
$('.dosurvey').click( function() {unloadPopupBox();});
}
}
$.cookie('cisit292014', $.cookie('cisit292014') + '1' , {expires: 15, path: '/'});
//alert($.cookie('cisit292014'));
}, 20000);
});
function dosurvey(pageURL, title, w, h) {
var left = (screen.width - w) / 2;
var top = (screen.height - h) / 6;
var targetWin = window.open(pageURL, title, 'toolbar=no, location=no, directories=no, status=yes, menubar=no, scrollbars=yes, resizable=yes, copyhistory=no, width=' + w + ', height=' + h + ', top=' + top + ', left=' + left);
}
</script>