离开网站时弹出 [英] Popup when leaving website
问题描述
我遇到了JavaScript问题. 我想要一个脚本，该脚本将在退出整个网站时弹出一个带有问题的消息，如果访问者回答否"，网页将关闭，如果访问者回答是"，则将被重定向到另一个页面.我在 http://www.pgrs.net上找到了一个示例/2008/01/30/popup-when-leaving-website/，但似乎对我不起作用.我找不到任何解决方案. 请检查我的代码，并告诉我也许我做错了什么? 这是我的源代码.
I got a problem with JavaScript. I want a script that will pop-up on exit whole web-site a message with question and if visitor answers "NO" web page closes and if he answers "YES" he will be redirected to another page. I found a example at http://www.pgrs.net/2008/01/30/popup-when-leaving-website/ but it seems that it doesnt work for me. I couldnt find any solution. Pleae check my code and tell me maybe i'm doing something wrong ? Here`s my source code.
也许有人会遇到问题.
Maybe somebody will see a problem.
<!DOCTYPE html>
<html lang="lt">
<head>
<meta charset="utf-8">
<title>PUA.LT</title>
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1">
<meta name="description" content="">
<meta name="author" content="Perfect WEB Solutions">
<link rel="stylesheet" href="<?php echo base_url("additional/style.css") ?>">
<script src='<?php echo base_url("additional/prototype.js")?>' type='text/javascript' ></script>
</head>
<body>
<script type="text/javascript">
Event.observe(document.body, 'click', function(event) {
if (Event.element(event).tagName == 'A') {
staying_in_site = true;
}
});
window.onunload = popup;
function popup() {
if(staying_in_site) {
return;
}
alert('I see you are leaving the site');
}
</script>
</body>
</html>
推荐答案
尝试一下:
window.onbeforeunload = popup;
function popup() {
return 'I see you are leaving the site';
}
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