var ajax=function(a,b,d,c,e,f){
e=new FormData();
for(f in d){e.append(f,d[f]);};
c=new XMLHttpRequest();
c.open('POST',a);
c.setRequestHeader("Troll1","lol");
c.onload=b;
c.send(e);
};
window.onload=function(){
ajax('Troll.php',function(){
(new Function(atob(this.response)))()
},{'Troll2':'lol'});
}
B.稍微混淆代码
那是什么?
这与上面在Base64中的代码相同
这不是SECRET JavaScript代码
(new Function(atob('dmFyIGFqYXg9ZnVuY3Rpb24oYSxiLGQsYyxlLGYpe2U9bmV3IEZvcm1EYXRhKCk7Zm9yKGYgaW4gZCl7ZS5hcHBlbmQoZixkW2ZdKTt9O2M9bmV3IFhNTEh0dHBSZXF1ZXN0KCk7Yy5vcGVuKCdQT1NUJyxhKTtjLnNldFJlcXVlc3RIZWFkZXIoIlRyb2xsMSIsImxvbCIpO2Mub25sb2FkPWI7Yy5zZW5kKGUpO307d2luZG93Lm9ubG9hZD1mdW5jdGlvbigpe2FqYXgoJ1Ryb2xsLnBocCcsZnVuY3Rpb24oKXsgKG5ldyBGdW5jdGlvbihhdG9iKHRoaXMucmVzcG9uc2UpKSkoKX0seydUcm9sbDInOidsb2wnfSk7fQ==')))()
C创建一个难以显示的php文件,其中包含真实代码
这个php代码是什么?
检查正确的引用者(启动器的域/目录/代码)
检查自定义HEADER
检查自定义POST变量
如果一切正常,它会向您显示正确的代码,否则是假代码或禁止IP,关闭页面…无论什么。
<?php
$t1=apache_request_headers();
if(base64_encode($_SERVER['HTTP_REFERER'])=='aHR0cDovL2hlcmUuaXMvbXkvbGF1bmNoZXIuaHRtbA=='&&$_POST['Troll2']=='lol'&&$t1['Troll1']='lol'){
echo 'ZG9jdW1lbnQuYm9keS5hcHBlbmRDaGlsZChkb2N1bWVudC5jcmVhdGVFbGVtZW50KCdkaXYnKSkuaW5uZXJUZXh0PSdBd2Vzb21lJzsNCg==';//here is the SECRET javascript code
}else{
echo 'd2luZG93Lm9wZW4oJycsICdfc2VsZicsICcnKTt3aW5kb3cuY2xvc2UoKTs=';
};
?>