有反正我可以使用JavaScript来阻止从另一个网站(IFRAME)弹出?如何防止使用JavaScript弹出?
例如:我有一个网站,其中几个其他网站的iframe。其中一人有这样的弹出脚本:
<script type="text/javascript">
var popunder=new Array()
popunder[0]="http://www.target.com"
//Specify the width and height of new popunder window (in pixels).
var width = '700';
var height = '450';
//these are obvious variables. set "yes" or "no".
var p = 'scrollbars=yes, resizable=yes, toolbar=yes,' + 'menubar=yes, status=yes, location=yes, left=85, top=20, height=' + height + ',width=' + width;
// Load new PopUnder only once per browser session? (0=no, 1=yes)
// Putting 0 will cause the Popunder to load every time page is loaded
// Specifying 1 will cause it to load only once per session
var one_time=0
// That's it! Don't edit the code below unless you're really good. :-P //
function get_cookie(Name) {
var search = Name + "="
var returnvalue = "";
if (document.cookie.length > 0) {
offset = document.cookie.indexOf(search)
if (offset != -1) { // if the cookie exists
offset += search.length
//set the index of beginning value
end = document.cookie.indexOf(";", offset);
if (end == -1) // set the index of the end of cookie value
end = document.cookie.length;
returnvalue = unescape(document.cookie.substring(offset, end))
}
}
return returnvalue;
}
function loadornot(){
if (get_cookie('popunder')==''){
load_pop_power()
document.cookie="popunder=yes"
}
}
function load_pop_power(){
win2 = window.open(popunder[Math.floor(Math.random()*(popunder.length))], "bw", p)
win2.blur()
window.focus()
}
if (one_time==0)
load_pop_power()
else
loadornot()
</script>
提供此弹出不能块和用户对Firefox或IE低安全设置。
我有ff。 iframe上的网站:(iframe.php)
<iframe src="http://friend.com/pop.php"></iframe>
iframe.php页面上应该怎么做以防止弹出?
我们可以XSS到网站吗?还是仅可以检测?我可以使用脚本删除iframe,如果检测到 – DucDigital 2011-04-18 05:33:37
对不起,我从来没有这样做,我甚至找不到解决方案,即使从javascript检测弹出。您可以在iframe中找到打开的模式窗口,但无法在JavaScript中找到通过iframe打开的正常窗口。唯一的方法来检测是使用浏览器插件! – 2011-04-18 06:49:19