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##缘起

在看资料时,看到这样的防止iframe嵌套的代码:

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try {
if (window.top != window.self) {
var ref = document.referer;
if (ref.substring(0, 2) === '//') {
ref = 'http:' + ref;
} else if (ref.split('://').length === 1) {
ref = 'http://' + ref;
}
var url = ref.split('/');
var _l = {auth: ''};
var host = url[2].split('@');
if (host.length === 1) {
host = host[0].split(':');
} else {
_l.auth = host[0];
host = host[1].split(':');
}
var parentHostName = host[0];
if (parentHostName.indexOf("test.com") == -1 && parentHostName.indexOf("test2.com") == -1) {
top.location.href = "http://www.test.com";
}
}
} catch (e) {
}

假定test.com,test2.com是自己的域名,当其它网站恶意嵌套本站的页面时,跳转回本站的首页。

上面的代码有两个问题:

  • referer拼写错误,实际上应该是referrer
  • 解析referrer的代码太复杂,还不一定正确

无论在任何语言里,都不建议手工写代码处理URL。因为url的复杂度超出一般人的想像。很多安全的问题就是因为解析URL不当引起的。比如防止CSRF时判断referrer。

URI的语法:

http://en.wikipedia.org/wiki/URI_scheme#Generic_syntax

##在javascript里解析url最好的办法

在javascript里解析url的最好办法是利用浏览器的js引擎,通过创建一个a标签:

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var getLocation = function(href) {
var l = document.createElement("a");
l.href = href;
return l;
};
var l = getLocation("http://example.com/path");
console.debug(l.hostname)

##简洁防iframe恶意嵌套的方法

下面给出一个简洁的防止iframe恶意嵌套的判断方法:

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if(window.top != window && document.referrer){
var a = document.createElement("a");
a.href = document.referrer;
var host = a.hostname;
var endsWith = function (str, suffix) {
return str.indexOf(suffix, str.length - suffix.length) !== -1;
}
if(!endsWith(host, '.test.com') || !endsWith(host, '.test2.com')){
top.location.href = "http://www.test.com";
}
}

##java里处理URL的方法
http://docs.oracle.com/javase/tutorial/networking/urls/urlInfo.html

用contain, indexOf, endWitch这些函数时都要小心。

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public static void main(String[] args) throws Exception {
URL aURL = new URL("http://example.com:80/docs/books/tutorial"
+ "/index.html?name=networking#DOWNLOADING");
System.out.println("protocol = " + aURL.getProtocol());
System.out.println("authority = " + aURL.getAuthority());
System.out.println("host = " + aURL.getHost());
System.out.println("port = " + aURL.getPort());
System.out.println("path = " + aURL.getPath());
System.out.println("query = " + aURL.getQuery());
System.out.println("filename = " + aURL.getFile());
System.out.println("ref = " + aURL.getRef());
}

##参考

http://stackoverflow.com/questions/736513/how-do-i-parse-a-url-into-hostname-and-path-in-javascript

http://stackoverflow.com/questions/5522097/prevent-iframe-stealing

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