0x01 問題提出
在一次演練中,我們通過wireshark抓取了一個如下的數據包,我們如何對其進行分析?
0x02 問題分析
流量包是如何捕獲的?
首先我們從上面的數據包分析可以知道,這是個USB的流量包,我們可以先嘗試分析一下USB的數據包是如何捕獲的。
在開始前,我們先介紹一些USB的基礎知識。USB有不同的規格,以下是使用USB的三種方式:
l USB UART
l USB HID
l USB Memory
UART或者Universal Asynchronous Receiver/Transmitter。這種方式下,設備只是簡單的將USB用於接受和發射數據,除此之外就再沒有其他通訊功能了。
HID是人性化的接口。這一類通訊適用於交互式,有這種功能的設備有:鍵盤,鼠標,游戲手柄和數字顯示設備。
最后是USB Memory,或者說是數據存儲。External HDD, thumb drive / flash drive,等都是這一類的。
其中使用的最廣的不是USB HID 就是USB Memory了。
每一個USB設備(尤其是HID或者Memory)都有一個供應商ID(Vendor Id)和產品識別碼(Product Id)。Vendor Id是用來標記哪個廠商生產了這個USB設備。Product Id用來標記不同的產品,他並不是一個特殊的數字,當然最好不同。如下圖
上圖是我在虛擬機環境下連接在我電腦上的USB設備列表,通過lsusb查看命令。
例如說,我在VMware下有一個無線鼠標。它是屬於HID設備。這個設備正常的運行,並且通過lsusb
這個命令查看所有USB設備,現在大家能找出哪一條是這個鼠標嗎??沒有錯,就是第四個,就是下面這條:
Bus 002 Device 002: ID 0e0f:0003 VMware, Inc. Virtual Mouse
其中,ID 0e0f:0003
就是Vendor-Product Id對,Vendor Id的值是0e0f,並且Product Id的值是0003。Bus 002 Device 002
代表usb設備正常連接,這點需要記下來。
我們用root權限運行Wireshark捕獲USB數據流。但是通常來說我們不建議這么做。我們需要給用戶足夠的權限來獲取linux中的usb數據流。我們可以用udev來達到我們的目的。我們需要創建一個用戶組usbmon,然后把我們的賬戶添加到這個組中。
addgroup usbmon gpasswd -a $USER usbmon echo 'SUBSYSTEM=="usbmon", GROUP="usbmon", MODE="640"' > /etc/udev/rules.d/99-usbmon.rules
接下來,我們需要usbmon內核模塊。如果該模塊沒有被加載,我們可以通過以下命令加載該模塊:
modprobe usbmon
打開wireshark,你會看到usbmonX其中X代表數字。下圖是我們本次的結果(我使用的是root):
如果接口處於活躍狀態或者有數據流經過的時候,wireshark的界面就會把它以波形圖的方式顯示出來。那么,我們該選那個呢?沒有錯,就是我剛剛讓大家記下來的,這個X的數字就是對應這USB Bus。在本文中是usbmon0。打開他就可以觀察數據包了。
通過這些,我們可以了解到usb設備與主機之間的通信過程和工作原理,我們可以來對流量包進行分析了。
如何去分析一個USB流量包?
根據前面的知識鋪墊,我們大致對USB流量包的抓取有了一個輪廓了,下面我們介紹一下如何分析一個USB流量包。
USB協議的細節方面參考wireshark的wiki:https://wiki.wireshark.org/USB
我們先拿GitHub上一個簡單的例子開始講起:
我們分析可以知道,USB協議的數據部分在Leftover Capture Data域之中,在Mac和Linux下可以用tshark命令可以將 leftover capture data單獨提取出來,命令如下:
tshark -r example.pcap -T fields -e usb.capdata //如果想導入usbdata.txt文件中,后面加上參數:>usbdata.txt
Windows下裝了wireshark的環境下,在wireshark目錄下有個tshark.exe,比如我的在D:\Program Files\Wireshark\tshark.exe
調用cmd,定位到當前目錄下,輸入如下命令即可:
tshark.exe -r example.pcap -T fields -e usb.capdata //如果想導入usbdata.txt文件中,后面加上參數:>usbdata.txt
有關tshark命令的詳細使用參考wireshark官方文檔:https://www.wireshark.org/docs/man-pages/tshark.html
運行命令並查看usbdata.txt發現數據包長度為八個字節
關於USB的特點應用我找了一張圖,很清楚的反應了這個問題:
這里我們只關注USB流量中的鍵盤流量和鼠標流量。
鍵盤數據包的數據長度為8個字節,擊鍵信息集中在第3個字節,每次key stroke都會產生一個keyboard event usb packet。
鼠標數據包的數據長度為4個字節,第一個字節代表按鍵,當取0x00時,代表沒有按鍵、為0x01時,代表按左鍵,為0x02時,代表當前按鍵為右鍵。第二個字節可以看成是一個signed byte類型,其最高位為符號位,當這個值為正時,代表鼠標水平右移多少像素,為負時,代表水平左移多少像素。第三個字節與第二字節類似,代表垂直上下移動的偏移。
我翻閱了大量的USB協議的文檔,在這里我們可以找到這個值與具體鍵位的對應關系:http://www.usb.org/developers/hidpage/Hut1_12v2.pdf
usb keyboard的映射表 根據這個映射表將第三個字節取出來,對應對照表得到解碼:
我們寫出如下腳本:
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)
結果如下:
我們把前面的整合成腳本,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:
另外貼上一份鼠標流量數據包轉換腳本:
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()
鍵盤流量數據包轉換腳本如下:
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
上面這個例子的項目鏈接如下:https://files.cnblogs.com/files/ECJTUACM-873284962/UsbKeyboardDataHacker.rar
那么對於我們開篇提到的問題,我們可以模仿嘗試如上這個例子:
首先我們通過tshark將usb.capdata全部導出:
tshark -r task_AutoKey.pcapng -T fields -e usb.capdata //如果想導入usbdata.txt文件中,后面加上參數:>usbdata.txt
我們用上面的python腳本將第三個字節取出來,對應對照表得到解碼:
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)
運行結果如下:
output :n[unknown]A[unknown]UTOKEY''.DECIPHER'[unknown]MPLRVFFCZEYOUJFJKYBXGZVDGQAURKXZOLKOLVTUFBLRNJESQITWAHXNSIJXPNMPLSHCJBTYHZEALOGVIAAISSPLFHLFSWFEHJNCRWHTINSMAMBVEXO[DEL]PZE[DEL]IZ'
我們可以看出這是自動密匙解碼,現在的問題是在我們不知道密鑰的情況下應該如何解碼呢?
我找到了如下這篇關於如何爆破密匙:http://www.practicalcryptography.com/cryptanalysis/stochastic-searching/cryptanalysis-autokey-cipher/
爆破腳本如下:
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)
跑出來的結果如下:
我們看到了flag的字樣,整理可得如下:
-674.914569565 autokey, klen 8 :"FLAGHERE", HELLOBOYSANDGIRLSYOUARESOSMARTTHATYOUCANFINDTHEFLAGTHATIHIDEINTHEKEYBOARDPACKAGEFLAGISJHAWLZKEWXHNCDHSLWBAQJTUQZDXZQPF
我們把字段進行分割看:
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 資源下載
本文涉及到的所有項目鏈接全部放在Github上:
0x04 擴展閱讀
- https://blog.csdn.net/songze_lee/article/details/77658094
- https://wiki.wireshark.org/USB
- http://www.usb.org/developers/hidpage/Hut1_12v2.pdf
- https://www.wireshark.org/docs/man-pages/tshark.html
- http://www.practicalcryptography.com/cryptanalysis/stochastic-searching/cryptanalysis-autokey-cipher/
- https://hackfun.org/2017/02/22/CTF%E4%B8%AD%E9%82%A3%E4%BA%9B%E8%84%91%E6%B4%9E%E5%A4%A7%E5%BC%80%E7%9A%84%E7%BC%96%E7%A0%81%E5%92%8C%E5%8A%A0%E5%AF%86/