Javascript 离开网站时弹出
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Popup when leaving website
提问by Daniil T.
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.
我遇到了 JavaScript 问题。我想要一个脚本，它会在退出整个网站时弹出一条带有问题的消息，如果访问者回答“否”网页关闭，如果他回答“是”，他将被重定向到另一个页面。我在http://www.pgrs.net/2008/01/30/popup-when-leaving-website/找到了一个例子，但它似乎对我不起作用。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。
Maybe somebody will see a problem.
也许有人会看到问题。
<!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>
回答by emphaticsunshine
Try this:
尝试这个：
window.onbeforeunload = popup;
function popup() {
return 'I see you are leaving the site';
}
