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防止页面被iframe恶意嵌套_html/css_WEB-ITnose

WBOY
WBOYOriginal
2016-06-24 11:48:391624browse

新blog地址:http://hengyunabc.github.io/prevent-iframe-stealing/

缘起

在看资料时,看到这样的防止iframe嵌套的代码:

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:

    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:

    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。

     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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