深入理解USB流量数据包的抓取与分析

0x01 问题提出

在一次演练中，我们通过wireshark抓取了一个如下的数据包，我们如何对其进行分析？

0x02 问题分析

流量包是如何捕获的？

首先我们从上面的数据包分析可以知道，这是个USB的流量包，我们可以先尝试分析一下USB的数据包是如何捕获的。

在开始前，我们先介绍一些USB的基础知识。USB有不同的规格，以下是使用USB的三种方式：

l USB UART
l USB HID
l USB Memory

UART或者Universal Asynchronous Receiver/Transmitter。这种方式下，设备只是简单的将USB用于接受和发射数据，除此之外就再没有其他通讯功能了。

HID是人性化的接口。这一类通讯适用于交互式，有这种功能的设备有：键盘，鼠标，游戏手柄和数字显示设备。

最后是USB Memory，或者说是数据存储。External HDD, thumb drive / flash drive,等都是这一类的。

其中使用的最广的不是USB HID 就是USB Memory了。

每一个USB设备（尤其是HID或者Memory）都有一个供应商ID（Vendor Id）和产品识别码（Product Id）。Vendor Id是用来标记哪个厂商生产了这个USB设备。Product Id用来标记不同的产品，他并不是一个特殊的数字，当然最好不同。如下图

上图是我在虚拟机环境下连接在我电脑上的USB设备列表，通过lsusb查看命令。

例如说，我在VMware下有一个无线鼠标。它是属于HID设备。这个设备正常的运行，并且通过lsusb这个命令查看所有USB设备，现在大家能找出哪一条是这个鼠标吗？？没有错，就是第四个，就是下面这条：

Bus 002 Device 002: ID 0e0f:0003 VMware, Inc. Virtual Mouse

其中，ID 0e0f:0003就是Vendor-Product Id对，Vendor Id的值是0e0f，并且Product Id的值是0003。Bus 002 Device 002代表usb设备正常连接，这点需要记下来。

我们用root权限运行Wireshark捕获USB数据流。但是通常来说我们不建议这么做。我们需要给用户足够的权限来获取linux中的usb数据流。我们可以用udev来达到我们的目的。我们需要创建一个用户组usbmon，然后把我们的账户添加到这个组中。

addgroup usbmon
gpasswd -a $USER usbmon
echo 'SUBSYSTEM=="usbmon", GROUP="usbmon", MODE="640"' > /etc/udev/rules.d/99-usbmon.rules

接下来，我们需要usbmon内核模块。如果该模块没有被加载，我们可以通过以下命令加载该模块：

modprobe usbmon

打开wireshark，你会看到usbmonX其中X代表数字。下图是我们本次的结果（我使用的是root）：

如果接口处于活跃状态或者有数据流经过的时候，wireshark的界面就会把它以波形图的方式显示出来。那么，我们该选那个呢？没有错，就是我刚刚让大家记下来的，这个X的数字就是对应这USB Bus。在本文中是usbmon0。打开他就可以观察数据包了。

通过这些，我们可以了解到usb设备与主机之间的通信过程和工作原理，我们可以来对流量包进行分析了。

如何去分析一个USB流量包？

根据前面的知识铺垫，我们大致对USB流量包的抓取有了一个轮廓了，下面我们介绍一下如何分析一个USB流量包。

USB协议的细节方面参考wireshark的wiki：https://wiki.wireshark.org/USB

我们先拿GitHub上一个简单的例子开始讲起：

我们分析可以知道，USB协议的数据部分在Leftover Capture Data域之中，在Mac和Linux下可以用tshark命令可以将 leftover capture data单独提取出来，命令如下：

tshark -r example.pcap -T fields -e usb.capdata //如果想导入usbdata.txt文件中，后面加上参数：>usbdata.txt

Windows下装了wireshark的环境下，在wireshark目录下有个tshark.exe，比如我的在D:\Program Files\Wireshark\tshark.exe

调用cmd，定位到当前目录下，输入如下命令即可：

tshark.exe -r example.pcap -T fields -e usb.capdata //如果想导入usbdata.txt文件中，后面加上参数：>usbdata.txt

有关tshark命令的详细使用参考wireshark官方文档：https://www.wireshark.org/docs/man-pages/tshark.html

运行命令并查看usbdata.txt发现数据包长度为八个字节

关于USB的特点应用我找了一张图，很清楚的反应了这个问题：

这里我们只关注USB流量中的键盘流量和鼠标流量。

键盘数据包的数据长度为8个字节，击键信息集中在第3个字节，每次key stroke都会产生一个keyboard event usb packet。

鼠标数据包的数据长度为4个字节，第一个字节代表按键，当取0x00时，代表没有按键、为0x01时，代表按左键，为0x02时，代表当前按键为右键。第二个字节可以看成是一个signed byte类型，其最高位为符号位，当这个值为正时，代表鼠标水平右移多少像素，为负时，代表水平左移多少像素。第三个字节与第二字节类似，代表垂直上下移动的偏移。

我翻阅了大量的USB协议的文档，在这里我们可以找到这个值与具体键位的对应关系：http://www.usb.org/developers/hidpage/Hut1_12v2.pdf

usb keyboard的映射表 根据这个映射表将第三个字节取出来，对应对照表得到解码：

我们写出如下脚本：

mappings = { 0x04:"A",  0x05:"B",  0x06:"C", 0x07:"D", 0x08:"E", 0x09:"F", 0x0A:"G",  0x0B:"H", 0x0C:"I",  0x0D:"J", 0x0E:"K", 0x0F:"L", 0x10:"M", 0x11:"N",0x12:"O",  0x13:"P", 0x14:"Q", 0x15:"R", 0x16:"S", 0x17:"T", 0x18:"U",0x19:"V", 0x1A:"W", 0x1B:"X", 0x1C:"Y", 0x1D:"Z", 0x1E:"1", 0x1F:"2", 0x20:"3", 0x21:"4", 0x22:"5",  0x23:"6", 0x24:"7", 0x25:"8", 0x26:"9", 0x27:"0", 0x28:"n", 0x2a:"[DEL]",  0X2B:"    ", 0x2C:" ",  0x2D:"-", 0x2E:"=", 0x2F:"[",  0x30:"]",  0x31:"\\", 0x32:"~", 0x33:";",  0x34:"'", 0x36:",",  0x37:"." }
nums = []
keys = open('usbdata.txt')
for line in keys:
    if line[0]!='0' or line[1]!='0' or line[3]!='0' or line[4]!='0' or line[9]!='0' or line[10]!='0' or line[12]!='0' or line[13]!='0' or line[15]!='0' or line[16]!='0' or line[18]!='0' or line[19]!='0' or line[21]!='0' or line[22]!='0':
         continue
    nums.append(int(line[6:8],16))
    # 00:00:xx:....
keys.close()
output = ""
for n in nums:
    if n == 0 :
        continue
    if n in mappings:
        output += mappings[n]
    else:
        output += '[unknown]'
print('output :n' + output)

结果如下：

我们把前面的整合成脚本，得：

#!/usr/bin/env python

import sys
import os

DataFileName = "usb.dat"

presses = []

normalKeys = {"04":"a", "05":"b", "06":"c", "07":"d", "08":"e", "09":"f", "0a":"g", "0b":"h", "0c":"i", "0d":"j", "0e":"k", "0f":"l", "10":"m", "11":"n", "12":"o", "13":"p", "14":"q", "15":"r", "16":"s", "17":"t", "18":"u", "19":"v", "1a":"w", "1b":"x", "1c":"y", "1d":"z","1e":"1", "1f":"2", "20":"3", "21":"4", "22":"5", "23":"6","24":"7","25":"8","26":"9","27":"0","28":"<RET>","29":"<ESC>","2a":"<DEL>", "2b":"\t","2c":"<SPACE>","2d":"-","2e":"=","2f":"[","30":"]","31":"\\","32":"<NON>","33":";","34":"'","35":"<GA>","36":",","37":".","38":"/","39":"<CAP>","3a":"<F1>","3b":"<F2>", "3c":"<F3>","3d":"<F4>","3e":"<F5>","3f":"<F6>","40":"<F7>","41":"<F8>","42":"<F9>","43":"<F10>","44":"<F11>","45":"<F12>"}

shiftKeys = {"04":"A", "05":"B", "06":"C", "07":"D", "08":"E", "09":"F", "0a":"G", "0b":"H", "0c":"I", "0d":"J", "0e":"K", "0f":"L", "10":"M", "11":"N", "12":"O", "13":"P", "14":"Q", "15":"R", "16":"S", "17":"T", "18":"U", "19":"V", "1a":"W", "1b":"X", "1c":"Y", "1d":"Z","1e":"!", "1f":"@", "20":"#", "21":"$", "22":"%", "23":"^","24":"&","25":"*","26":"(","27":")","28":"<RET>","29":"<ESC>","2a":"<DEL>", "2b":"\t","2c":"<SPACE>","2d":"_","2e":"+","2f":"{","30":"}","31":"|","32":"<NON>","33":"\"","34":":","35":"<GA>","36":"<","37":">","38":"?","39":"<CAP>","3a":"<F1>","3b":"<F2>", "3c":"<F3>","3d":"<F4>","3e":"<F5>","3f":"<F6>","40":"<F7>","41":"<F8>","42":"<F9>","43":"<F10>","44":"<F11>","45":"<F12>"}

def main():
    # check argv
    if len(sys.argv) != 2:
        print "Usage : "
        print "        python UsbKeyboardHacker.py data.pcap"
        print "Tips : "
        print "        To use this python script , you must install the tshark first."
        print "        You can use `sudo apt-get install tshark` to install it"
        print "Author : "
        print "        Angel_Kitty <angelkitty6698@gmail.com>"
        print "        If you have any questions , please contact me by email."
        print "        Thank you for using."
        exit(1)

    # get argv
    pcapFilePath = sys.argv[1]
    
    # get data of pcap
    os.system("tshark -r %s -T fields -e usb.capdata > %s" % (pcapFilePath, DataFileName))

    # read data
    with open(DataFileName, "r") as f:
        for line in f:
            presses.append(line[0:-1])
    # handle
    result = ""
    for press in presses:
        Bytes = press.split(":")
        if Bytes[0] == "00":
            if Bytes[2] != "00":
                result += normalKeys[Bytes[2]]
        elif Bytes[0] == "20": # shift key is pressed.
            if Bytes[2] != "00":
                result += shiftKeys[Bytes[2]]
        else:
            print "[-] Unknow Key : %s" % (Bytes[0])
    print "[+] Found : %s" % (result)

    # clean the temp data
    os.system("rm ./%s" % (DataFileName))


if __name__ == "__main__":
    main()

效果如下：

另外贴上一份鼠标流量数据包转换脚本：

nums = [] 
keys = open('usbdata.txt','r') 
posx = 0 
posy = 0 
for line in keys: 
if len(line) != 12 : 
     continue 
x = int(line[3:5],16) 
y = int(line[6:8],16) 
if x > 127 : 
    x -= 256 
if y > 127 : 
    y -= 256 
posx += x 
posy += y 
btn_flag = int(line[0:2],16)  # 1 for left , 2 for right , 0 for nothing 
if btn_flag == 1 : 
    print posx , posy 
keys.close()

键盘流量数据包转换脚本如下：

nums=[0x66,0x30,0x39,0x65,0x35,0x34,0x63,0x31,0x62,0x61,0x64,0x32,0x78,0x33,0x38,0x6d,0x76,0x79,0x67,0x37,0x77,0x7a,0x6c,0x73,0x75,0x68,0x6b,0x69,0x6a,0x6e,0x6f,0x70]
s=''
for x in nums:
    s+=chr(x)
print s
mappings = { 0x41:"A",  0x42:"B",  0x43:"C", 0x44:"D", 0x45:"E", 0x46:"F", 0x47:"G",  0x48:"H", 0x49:"I",  0x4a:"J", 0x4b:"K", 0x4c:"L", 0x4d:"M", 0x4e:"N",0x4f:"O",  0x50:"P", 0x51:"Q", 0x52:"R", 0x53:"S", 0x54:"T", 0x55:"U",0x56:"V", 0x57:"W", 0x58:"X", 0x59:"Y", 0x5a:"Z", 0x60:"0", 0x61:"1", 0x62:"2", 0x63:"3", 0x64:"4",  0x65:"5", 0x66:"6", 0x67:"7", 0x68:"8", 0x69:"9", 0x6a:"*", 0x6b:"+",  0X6c:"separator", 0x6d:"-",  0x6e:".", 0x6f:"/" }
output = ""
for n in nums:
    if n == 0 :
        continue
    if n in mappings:
        output += mappings[n]
    else:
        output += '[unknown]'
print 'output :\n' + output

上面这个例子的项目链接如下：https://files.cnblogs.com/files/ECJTUACM-873284962/UsbKeyboardDataHacker.rar

那么对于我们开篇提到的问题，我们可以模仿尝试如上这个例子：

首先我们通过tshark将usb.capdata全部导出：

tshark -r task_AutoKey.pcapng -T fields -e usb.capdata //如果想导入usbdata.txt文件中，后面加上参数：>usbdata.txt

我们用上面的python脚本将第三个字节取出来，对应对照表得到解码：

mappings = { 0x04:"A",  0x05:"B",  0x06:"C", 0x07:"D", 0x08:"E", 0x09:"F", 0x0A:"G",  0x0B:"H", 0x0C:"I",  0x0D:"J", 0x0E:"K", 0x0F:"L", 0x10:"M", 0x11:"N",0x12:"O",  0x13:"P", 0x14:"Q", 0x15:"R", 0x16:"S", 0x17:"T", 0x18:"U",0x19:"V", 0x1A:"W", 0x1B:"X", 0x1C:"Y", 0x1D:"Z", 0x1E:"1", 0x1F:"2", 0x20:"3", 0x21:"4", 0x22:"5",  0x23:"6", 0x24:"7", 0x25:"8", 0x26:"9", 0x27:"0", 0x28:"n", 0x2a:"[DEL]",  0X2B:"    ", 0x2C:" ",  0x2D:"-", 0x2E:"=", 0x2F:"[",  0x30:"]",  0x31:"\\", 0x32:"~", 0x33:";",  0x34:"'", 0x36:",",  0x37:"." }
nums = []
keys = open('usbdata.txt')
for line in keys:
    if line[0]!='0' or line[1]!='0' or line[3]!='0' or line[4]!='0' or line[9]!='0' or line[10]!='0' or line[12]!='0' or line[13]!='0' or line[15]!='0' or line[16]!='0' or line[18]!='0' or line[19]!='0' or line[21]!='0' or line[22]!='0':
         continue
    nums.append(int(line[6:8],16))
    # 00:00:xx:....
keys.close()
output = ""
for n in nums:
    if n == 0 :
        continue
    if n in mappings:
        output += mappings[n]
    else:
        output += '[unknown]'
print('output :n' + output)

运行结果如下：

output :n[unknown]A[unknown]UTOKEY''.DECIPHER'[unknown]MPLRVFFCZEYOUJFJKYBXGZVDGQAURKXZOLKOLVTUFBLRNJESQITWAHXNSIJXPNMPLSHCJBTYHZEALOGVIAAISSPLFHLFSWFEHJNCRWHTINSMAMBVEXO[DEL]PZE[DEL]IZ'

我们可以看出这是自动密匙解码，现在的问题是在我们不知道密钥的情况下应该如何解码呢？

我找到了如下这篇关于如何爆破密匙：http://www.practicalcryptography.com/cryptanalysis/stochastic-searching/cryptanalysis-autokey-cipher/

爆破脚本如下：

from ngram_score import ngram_score
from pycipher import Autokey
import re
from itertools import permutations

qgram = ngram_score('quadgrams.txt')
trigram = ngram_score('trigrams.txt')

ctext = 'MPLRVFFCZEYOUJFJKYBXGZVDGQAURKXZOLKOLVTUFBLRNJESQITWAHXNSIJXPNMPLSHCJBTYHZEALOGVIAAISSPLFHLFSWFEHJNCRWHTINSMAMBVEXPZIZ'

ctext = re.sub(r'[^A-Z]','',ctext.upper())
# keep a list of the N best things we have seen, discard anything else

class nbest(object):
    def __init__(self,N=1000):
        self.store = []
        self.N = N

    def add(self,item):
        self.store.append(item)
        self.store.sort(reverse=True)
        self.store = self.store[:self.N]

    def __getitem__(self,k):
        return self.store[k]

    def __len__(self):
        return len(self.store)
        
#init
N=100
for KLEN in range(3,20):
    rec = nbest(N)
    for i in permutations('ABCDEFGHIJKLMNOPQRSTUVWXYZ',3):
        key = ''.join(i) + 'A'*(KLEN-len(i))
        pt = Autokey(key).decipher(ctext)
        score = 0
        for j in range(0,len(ctext),KLEN):
            score += trigram.score(pt[j:j+3])
        rec.add((score,''.join(i),pt[:30]))

    next_rec = nbest(N)
    for i in range(0,KLEN-3):
        for k in xrange(N):
            for c in 'ABCDEFGHIJKLMNOPQRSTUVWXYZ':
                key = rec[k][1] + c
                fullkey = key + 'A'*(KLEN-len(key))
                pt = Autokey(fullkey).decipher(ctext)
                score = 0
                for j in range(0,len(ctext),KLEN):
                    score += qgram.score(pt[j:j+len(key)])
                next_rec.add((score,key,pt[:30]))
        rec = next_rec
        next_rec = nbest(N)
    bestkey = rec[0][1]
    pt = Autokey(bestkey).decipher(ctext)
    bestscore = qgram.score(pt)
    for i in range(N):
        pt = Autokey(rec[i][1]).decipher(ctext)
        score = qgram.score(pt)
        if score > bestscore:
            bestkey = rec[i][1]
            bestscore = score       
    print bestscore,'autokey, klen',KLEN,':"'+bestkey+'",',Autokey(bestkey).decipher(ctext)

跑出来的结果如下：

我们看到了flag的字样，整理可得如下：

-674.914569565 autokey, klen 8 :"FLAGHERE", HELLOBOYSANDGIRLSYOUARESOSMARTTHATYOUCANFINDTHEFLAGTHATIHIDEINTHEKEYBOARDPACKAGEFLAGISJHAWLZKEWXHNCDHSLWBAQJTUQZDXZQPF

我们把字段进行分割看：

HELLO
BOYS
AND
GIRLS
YOU
ARE
SO
SMART
THAT
YOU
CAN
FIND
THE
FLAG
THAT
IH
IDE
IN
THE
KEY
BOARD
PACKAGE
FLAG
IS
JHAWLZKEWXHNCDHSLWBAQJTUQZDXZQPF

最后的flag就是flag{JHAWLZKEWXHNCDHSLWBAQJTUQZDXZQPF}

0x03 资源下载

本文涉及到的所有项目链接全部放在Github上：

0x04 扩展阅读

posted @ 2018-08-14 12:03  Angel_Kitty  阅读(18986)  评论(3)  编辑  收藏  举报