OverTheWire:Bandit通关WriteUp（2019.01.17完）
OverTheWire:Bandit通關全攻略WriteUp
- 背景
- 通關過程
- Level 0
- Level 0-->Level 1
- Level 1 - Level 2
- Level 2 - Level 3
- Level 3 → Level 4
- Level 4 → Level 5
- Level 5 → Level 6
- Level 6 → Level 7
- Level 7 → Level 8
- Level 8 → Level 9
- Level 9 → Level 10
- Level 10 → Level 11
- Level 11 → Level 12
- Level 12 → Level 13
- Level 13 → Level 14
- Level 14 → Level 15
- Level 15 → Level 16
- Level 16 → Level 17
- Level 17 → Level 18
- Level 18 → Level 19
- Level 19 → Level 20
- Level 20 → Level 21
- Level 21 → Level 22
- Level 22 → Level 23
- Level 23 → Level 24
- Level 24 → Level 25
- Level 25 → Level 26
- Level 26 → Level 27
- Level 27 → Level 28
- Level 28 → Level 29
- Level 29 → Level 30
- Level 30 → Level 31
- Level 31 → Level 32
- Level 32 → Level 33
背景
OverTheWire:Bandit是一個學習linux命令的WarGame，通過闖關的模式，不斷的學習新的命令，對于學習安全和Linux的朋友是一個很好的練習游戲，網址是 http://overthewire.org/wargames/bandit/ 。
這個游戲目前有34關，從Level0—Level34。游戲形式是通過ssh連接游戲服務器，通過各種命令行讀取下一關的游戲服務器密鑰，然后連接下一關的服務器繼續(xù)讀取，直到通關。
通關過程
Level 0
The goal of this level is for you to log into the game using SSH. The host to which you need to connect is bandit.labs.overthewire.org, on port 2220. The username is bandit0 and the password is bandit0. Once logged in, go to the Level 1 page to find out how to beat Level 1.
這一關主要是讓你選擇一個合適ssh工具開始遠程，這一關的用戶名和密碼均為bandit0常見的有secureCRT，Xshell, Putty, 不過我最近發(fā)現(xiàn)一款免費而且不比Xshell功能少的SSH工具叫MobaXterm，個人推薦。
Linux下更為方便
Level 0–>Level 1
he password for the next level is stored in a file called readme located in the home directory. Use this password to log into bandit1 using SSH. Whenever you find a password for a level, use SSH (on port 2220) to log into that level and continue the game.
Commands you may need to solve this level
ls, cd, cat, file, du, find
其中du命令是用來查看令也是查看使用空間的，但是與df命令不同的是Linux du命令是查看當前指定文件或目錄(會遞歸顯示子目錄)占用磁盤空間大小，還是和df命令有一些區(qū)別的
得到下一關用戶名bandit1，密碼為boJ9jbbUNNfktd78OOpsqOltutMc3MY1，之后用戶名依次類推，不做贅述
Level 1 - Level 2
Level Goal
The password for the next level is stored in a file called - located in the home directory
Commands you may need to solve this level
ls, cd, cat, file, du, find
ls發(fā)現(xiàn)文件名是一個-，但是這個在linux中有特殊意義導致直接cat不好用
直接輸入絕對路徑讀取
Level 2 - Level 3
Level Goal
The password for the next level is stored in a file called spaces in this filename located in the home directory
Commands you may need to solve this level
ls, cd, cat, file, du, find
Helpful Reading Material
Google Search for “spaces in filename”
文件名有空格的讀取
用cat命令，然后Tab按鍵補齊，自動將空格轉義，實現(xiàn)了密鑰讀取，或者給文件名加上雙引號也可以讀取。
Level 3 → Level 4
Level Goal
The password for the next level is stored in a hidden file in the inhere directory.
Commands you may need to solve this level
ls, cd, cat, file, du, find
密鑰寫在一個隱藏文件里面，通過ls -a參數(shù)可以找到隱藏文件
Level 4 → Level 5
Level Goal
The password for the next level is stored in the only human-readable file in the inhere directory. Tip: if your terminal is messed up, try the “reset” command.
Commands you may need to solve this level
ls, cd, cat, file, du, find
文件說在人類能讀懂的文件里面，可以看到當前目錄有9個文件，通過file命令可以用于辨識文件類型。
Level 5 → Level 6
Level Goal
The password for the next level is stored in a file somewhere under the inhere directory and has all of the following properties:
human-readable
1033 bytes in size
not executable
Commands you may need to solve this level
ls, cd, cat, file, du, find
一看有這么多文件夾
bandit5@bandit:~$ ls inhere bandit5@bandit:~$ cd inhere/ bandit5@bandit:~/inhere$ ls -a . maybehere02 maybehere06 maybehere10 maybehere14 maybehere18 .. maybehere03 maybehere07 maybehere11 maybehere15 maybehere19 maybehere00 maybehere04 maybehere08 maybehere12 maybehere16 maybehere01 maybehere05 maybehere09 maybehere13 maybehere17根據特征我們可以用find 命令，找到一個符合條件的文件
bandit5@bandit:~/inhere$ find . -type f -size 1033c ./maybehere07/.file2 bandit5@bandit:~/inhere$ cat ./maybehere07/.file2 DXjZPULLxYr17uwoI01bNLQbtFemEgo7附find參數(shù)解析
-size n[cwbkMG] : 檔案大小 為 n 個由后綴決定的數(shù)據塊。其中后綴含義為：
b: 代表 512 位元組的區(qū)塊（如果用戶沒有指定后綴，則默認為 b）
c: 表示字節(jié)數(shù)
k: 表示 kilo bytes （1024字節(jié)）
w: 字 （2字節(jié)）
M:兆字節(jié)（1048576字節(jié)）
G: 千兆字節(jié) （1073741824字節(jié)）
-type c : 檔案類型是 c 。
d: 目錄
c: 字型裝置檔案
b: 區(qū)塊裝置檔案
p: 具名貯列
f: 一般檔案
l: 符號連結
s: socket
Level 6 → Level 7
Level Goal
The password for the next level is stored somewhere on the server and has all of the following properties:
owned by user bandit7
owned by group bandit6
33 bytes in size
Commands you may need to solve this level
ls, cd, cat, file, du, find, grep
又是找文件，那么依然可以使用find命令，只不過參數(shù)稍稍的改變
后面的2>/dev/null因為find命令在根目錄下查找會經常有很多權限的報錯信息，所有在linux中通常用這種方式將錯誤信息重定向到“黑洞中”
Level 7 → Level 8
Level Goal
The password for the next level is stored in the file data.txt next to the word millionth
Commands you may need to solve this level
grep, sort, uniq, strings, base64, tr, tar, gzip, bzip2, xxd
根據提示data.txt中在密鑰在millionth中，可以通過grep命令查看
bandit7@bandit:~$ ls data.txt bandit7@bandit:~$ cat data.txt |grep millionth millionth cvX2JJa4CFALtqS87jk27qwqGhBM9plVLevel 8 → Level 9
Level Goal
The password for the next level is stored in the file data.txt and is the only line of text that occurs only once
Commands you may need to solve this level
grep, sort, uniq, strings, base64, tr, tar, gzip, bzip2, xxd
Helpful Reading Material
The unix commandline: pipes and redirects
這題是要找到出現(xiàn)一次的那個行，肯定用uniq命令了，但是使用之前需要用sort命令對文本進行排序，因為uniq命令是通過判斷上下兩行是否一樣來判斷的，所以用sort排序一下然后在uniq就能找到唯一出現(xiàn)的那一行了
sort data.txt|uniq -u
sort data.txt|uniq -c
這題找到兩種解法，一個是直接-u獲取，還有就是-c列出出現(xiàn)的次數(shù)，然后從中找到是1的那一行即可
Level 9 → Level 10
Level Goal
The password for the next level is stored in the file data.txt in one of the few human-readable strings, beginning with several ‘=’ characters.
Commands you may need to solve this level
grep, sort, uniq, strings, base64, tr, tar, gzip, bzip2, xxd
直接使用cat命令是很多很雜亂的東西，可以通過string命令查看文件中的字符串，根據提示信息可得下一關密鑰以若干個“=”開頭，可以找到下一關的密鑰truKLdjsbJ5g7yyJ2X2R0o3a5HQJFuLk
bandit9@bandit:~$ ls data.txt bandit9@bandit:~$ strings data.txt .MBB `B6ha t8lHX u ========== password NHGu 5xhH ========== truKLdjsbJ5g7yyJ2X2R0o3a5HQJFuLk W.uO7 i$2w epg~Level 10 → Level 11
Level Goal
The password for the next level is stored in the file data.txt, which contains base64 encoded data
Commands you may need to solve this level
grep, sort, uniq, strings, base64, tr, tar, gzip, bzip2, xxd
Helpful Reading Material
Base64 on Wikipedia
題目提示密鑰信息用了base64解碼，我們解碼即可
Level 11 → Level 12
Level Goal
The password for the next level is stored in the file data.txt, where all lowercase (a-z) and uppercase (A-Z) letters have been rotated by 13 positions
Commands you may need to solve this level
grep, sort, uniq, strings, base64, tr, tar, gzip, bzip2, xxd
Helpful Reading Material
Rot13 on Wikipedia
tr用來從標準輸入中通過替換或刪除操作進行字符轉換。tr主要用于刪除文件中控制字符或進行字符轉換。使用tr時要轉換兩個字符串：字符串1用于查詢，字符串2用于處理各種轉換。tr剛執(zhí)行時，字符串1中的字符被映射到字符串2中的字符，然后轉換操作開始。
帶有最常用選項的tr命令格式為：
tr -c -d -s [“string1_to_translate_from”][“string2_to_translate_to”] < input-file
Rot13是一種特殊的凱撒密碼轉換，根據題目所說的字母的的順序旋轉了13個位置，就相當去26個字母的前13個位置與后13個位置調換了。那么我們就是用tr命令進行調換
bandit11@bandit:~$ cat data.txt Gur cnffjbeq vf 5Gr8L4qetPEsPk8htqjhRK8XSP6x2RHh bandit11@bandit:~$ cat data.txt |tr 'a-zA-Z' 'n-za-mN-ZA-M' The password is 5Te8Y4drgCRfCx8ugdwuEX8KFC6k2EUuLevel 12 → Level 13
Level Goal
The password for the next level is stored in the file data.txt, which is a hexdump of a file that has been repeatedly compressed. For this level it may be useful to create a directory under /tmp in which you can work using mkdir. For example: mkdir /tmp/myname123. Then copy the datafile using cp, and rename it using mv (read the manpages!)
Commands you may need to solve this level
grep, sort, uniq, strings, base64, tr, tar, gzip, bzip2, xxd, mkdir, cp, mv
Helpful Reading Material
Hex dump on Wikipedia
這是一道比較麻煩的題目，需要我們解壓很多層。
可以看到這本來是一個文本類型的文件，嘗試用xxd轉成bin提示權限不夠，我們先復制一遍。
bandit12@bandit:~$ mkdir /tmp/c1911 bandit12@bandit:~$ cp data.txt /tmp/c1911 bandit12@bandit:~$ bandit12@bandit:~$ cd /tmp/c1911 bandit12@bandit:/tmp/c1911$ ls data.txt bandit12@bandit:/tmp/c1911$ xxd -r data.txt > data.bin bandit12@bandit:/tmp/c1911$ file data.bin data.bin: gzip compressed data, was "data2.bin", last modified: Tue Oct 16 12:00:23 2018, max compression, from Unix復制完是gzip格式，改文件名，解壓。
bandit12@bandit:/tmp/c1911$ mv data.bin data.gz bandit12@bandit:/tmp/c1911$ gzip -d data.gz bandit12@bandit:/tmp/c1911$ ls data data.txt bandit12@bandit:/tmp/c1911$ file data data: bzip2 compressed data, block size = 900k還有一層bzip2 ???, 繼續(xù)解壓
bandit12@bandit:/tmp/c1911$ mv data data.bz2 bandit12@bandit:/tmp/c1911$ bunzip2 -d data.bz2 bandit12@bandit:/tmp/c1911$ file data data: gzip compressed data, was "data4.bin", last modified: Tue Oct 16 12:00:23 2018, max compression, from Unix還有沒有解壓的文件，繼續(xù)搞搞吧！！！一直一直查看文件類型，重命名，解壓。直到第八層壓縮。。。。。
bandit12@bandit:/tmp/c1911$ mv data data.gz bandit12@bandit:/tmp/c1911$ gzip -d data.gz bandit12@bandit:/tmp/c1911$ file data data: POSIX tar archive (GNU) bandit12@bandit:/tmp/c1911$ mv data data.tar bandit12@bandit:/tmp/c1911$ tar xvf data.tar data5.bin bandit12@bandit:/tmp/c1911$ file data5.bin data5.bin: POSIX tar archive (GNU) bandit12@bandit:/tmp/c1911$ mv data5.bin data5.tar bandit12@bandit:/tmp/c1911$ tar xvf data5.tar data6.bin bandit12@bandit:/tmp/c1911$ file data6 data6: cannot open `data6' (No such file or directory) bandit12@bandit:/tmp/c1911$ file data6.bin data6.bin: bzip2 compressed data, block size = 900k bandit12@bandit:/tmp/c1911$ bunzip2 -d data6.bin bunzip2: Can't guess original name for data6.bin -- using data6.bin.out bandit12@bandit:/tmp/c1911$ file data6.bin.out data6.bin.out: POSIX tar archive (GNU) bandit12@bandit:/tmp/c1911$ mv data6.bin.out data.tar bandit12@bandit:/tmp/c1911$ tar xvf data.tar data8.bin bandit12@bandit:/tmp/c1911$ file data8.bin data8.bin: gzip compressed data, was "data9.bin", last modified: Tue Oct 16 12:00:23 2018, max compression, from Unix bandit12@bandit:/tmp/c1911$ mv data8.bin data8.gz bandit12@bandit:/tmp/c1911$ gzip -d data8.gz bandit12@bandit:/tmp/c1911$ ls data5.tar data8 data.tar data.txt bandit12@bandit:/tmp/c1911$ cat data8 The password is 8ZjyCRiBWFYkneahHwxCv3wb2a1ORpYLLevel 13 → Level 14
Level Goal
The password for the next level is stored in /etc/bandit_pass/bandit14 and can only be read by user bandit14. For this level, you don’t get the next password, but you get a private SSH key that can be used to log into the next level. Note: localhost is a hostname that refers to the machine you are working on
Commands you may need to solve this level
ssh, telnet, nc, openssl, s_client, nmap
Helpful Reading Material
SSH/OpenSSH/Keys
這一關告訴我們下一關的密碼存放在/etc目錄下，且只有bandit14用戶可讀，我們當前目錄下只有一個私鑰文件，可以考慮用私鑰文件去連接bandit14, 用bandit14 讀取用戶文件。
Level 14 → Level 15
Level Goal
The password for the next level can be retrieved by submitting the password of the current level to port 30000 on localhost.
Commands you may need to solve this level
ssh, telnet, nc, openssl, s_client, nmap
Helpful Reading Material
How the Internet works in 5 minutes (YouTube) (Not completely accurate, but good enough for beginners)
IP Addresses
IP Address on Wikipedia
Localhost on Wikipedia
Ports
Port (computer networking) on Wikipedia
這關說只要把本關的密鑰提交即可得到反饋，看來我直接從bandit13 ssh連接到的bandit14 可以說并不是算過了這一關，還是要拿到這一關的密鑰信息才能進行下一關，這也是這個游戲設計的一個巧妙之處吧。
Level 15 → Level 16
Level Goal
The password for the next level can be retrieved by submitting the password of the current level to port 30001 on localhost using SSL encryption.
Helpful note: Getting “HEARTBEATING” and “Read R BLOCK”? Use -ign_eof and read the “CONNECTED COMMANDS” section in the manpage. Next to ‘R’ and ‘Q’, the ‘B’ command also works in this version of that command…
Commands you may need to solve this level
ssh, telnet, nc, openssl, s_client, nmap
Helpful Reading Material
Secure Socket Layer/Transport Layer Security on Wikipedia
OpenSSL Cookbook - Testing with OpenSSL
這題說是要通過ssl發(fā)送本關密碼才可以的獲得下一關的密鑰信息。需要用到openssl。
Level 16 → Level 17
Level Goal
The credentials for the next level can be retrieved by submitting the password of the current level to a port on localhost in the range 31000 to 32000. First find out which of these ports have a server listening on them. Then find out which of those speak SSL and which don’t. There is only 1 server that will give the next credentials, the others will simply send back to you whatever you send to it.
Commands you may need to solve this level
ssh, telnet, nc, openssl, s_client, nmap
Helpful Reading Material
Port scanner on Wikipedia
這一題說開放的端口在31000和32000中間的某一個開放了ssl服務的端口上，肯定要使用到端口掃描程序，這里我們就使用nmap, 掃描一個端口范圍，找到我們應該使用的端口號
可以看到31518端口和31790端口開放了ssl服務，我們繼續(xù)連接這個端口發(fā)送本關密鑰。發(fā)現(xiàn)31518端口會將我們發(fā)送的內容直接返回，31790才是返回密碼的正確端口。
bandit16@bandit:~$ openssl s_client -connect localhost -port 31790 CONNECTED(00000003) depth=0 CN = localhost verify error:num=18:self signed certificate verify return:1 depth=0 CN = localhost verify return:1 --- Certificate chain0 s:/CN=localhosti:/CN=localhost --- Server certificate -----BEGIN CERTIFICATE----- MIICBjCCAW+gAwIBAgIENT6X8jANBgkqhkiG9w0BAQUFADAUMRIwEAYDVQQDDAls b2NhbGhvc3QwHhcNMTgxMjExMTAwMTQyWhcNMTkxMjExMTAwMTQyWjAUMRIwEAYD VQQDDAlsb2NhbGhvc3QwgZ8wDQYJKoZIhvcNAQEBBQADgY0AMIGJAoGBAJuYYSnx pA49LOi31RUGpW+JNJvStNuBSiMx17bhMuNlijN+b19LuSV1mWOAmo+zzIsBq5Yv CbvXfCKrjJnxEGuP+XtPmC3trplmej2jlRa/sRmKDIuV74Ze0GjzO25TY6a5XW+J lC0fqLCH/YsculqmLp8atEYYSaduS5vrz8ILAgMBAAGjZTBjMBQGA1UdEQQNMAuC CWxvY2FsaG9zdDBLBglghkgBhvhCAQ0EPhY8QXV0b21hdGljYWxseSBnZW5lcmF0 ZWQgYnkgTmNhdC4gU2VlIGh0dHBzOi8vbm1hcC5vcmcvbmNhdC8uMA0GCSqGSIb3 DQEBBQUAA4GBAACqyQVna9ckIFWR3EzUKXl7JgkCN0BK9Wy2rqzgPiplvuhjd4lC mlIDeEy7VnZoOEXj+YX4OIhnajkXi6XfIpBYP+/RPo27n5PgBT3ywJkWwVKWgaPa BFH7qSw6FBJZkiN5i25FSYXdg4/JT+/C6SZxDy5YGKLFoA3dEGCZ8hh7 -----END CERTIFICATE----- subject=/CN=localhost issuer=/CN=localhost --- No client certificate CA names sent Peer signing digest: SHA512 Server Temp Key: X25519, 253 bits --- SSL handshake has read 1019 bytes and written 269 bytes Verification error: self signed certificate --- New, TLSv1.2, Cipher is ECDHE-RSA-AES256-GCM-SHA384 Server public key is 1024 bit Secure Renegotiation IS supported Compression: NONE Expansion: NONE No ALPN negotiated SSL-Session:Protocol : TLSv1.2Cipher : ECDHE-RSA-AES256-GCM-SHA384Session-ID: FC46E2669B162F04B5C370807C1B9E92FCEA3B123059C3E4701A30C1E749B661Session-ID-ctx:Master-Key: 54B4F5C34BE5CE0F33249917300889499881A634B2D13715698130E69E07A1F92B55E3AE86074A7ED3E0DAE14264F3DDPSK identity: NonePSK identity hint: NoneSRP username: NoneTLS session ticket lifetime hint: 7200 (seconds)TLS session ticket:0000 - 2e 71 2f 27 00 eb 72 19-5f 50 c2 d0 8e 8e 6f 16 .q/'..r._P....o.0010 - f3 18 94 00 14 11 41 35-a2 b9 c9 d9 d1 a3 87 cb ......A5........0020 - d1 d5 9c 82 4e 31 5a e2-ec 49 a3 1e 37 eb 8d fe ....N1Z..I..7...0030 - c9 ce cc c0 72 26 b8 42-70 86 71 5e 0a d6 35 77 ....r&.Bp.q^..5w0040 - 3d 49 07 54 d7 e7 17 d1-b6 20 0d 9c 62 bf 7c db =I.T..... ..b.|.0050 - d3 a5 de bf 89 33 f4 c3-21 b1 88 7b dc 3b e4 11 .....3..!..{.;..0060 - fe 0a 43 d0 62 c2 b2 7c-94 62 cf 98 a3 b3 64 5e ..C.b..|.b....d^0070 - 70 c1 9f fe 6d 2f 2d 40-36 6c f6 79 72 cb 30 d0 p...m/-@6l.yr.0.0080 - 89 e8 f9 83 5b 7d 65 0d-b2 ed 17 68 ad ae 6b 68 ....[}e....h..kh0090 - ad 5f ce 31 7f b3 ec e0-36 c9 f0 e6 de 0c 24 9a ._.1....6.....$.Start Time: 1546576890Timeout : 7200 (sec)Verify return code: 18 (self signed certificate)Extended master secret: yes --- cluFn7wTiGryunymYOu4RcffSxQluehd Correct! -----BEGIN RSA PRIVATE KEY----- MIIEogIBAAKCAQEAvmOkuifmMg6HL2YPIOjon6iWfbp7c3jx34YkYWqUH57SUdyJ imZzeyGC0gtZPGujUSxiJSWI/oTqexh+cAMTSMlOJf7+BrJObArnxd9Y7YT2bRPQ Ja6Lzb558YW3FZl87ORiO+rW4LCDCNd2lUvLE/GL2GWyuKN0K5iCd5TbtJzEkQTu DSt2mcNn4rhAL+JFr56o4T6z8WWAW18BR6yGrMq7Q/kALHYW3OekePQAzL0VUYbW JGTi65CxbCnzc/w4+mqQyvmzpWtMAzJTzAzQxNbkR2MBGySxDLrjg0LWN6sK7wNX x0YVztz/zbIkPjfkU1jHS+9EbVNj+D1XFOJuaQIDAQABAoIBABagpxpM1aoLWfvD KHcj10nqcoBc4oE11aFYQwik7xfW+24pRNuDE6SFthOar69jp5RlLwD1NhPx3iBl J9nOM8OJ0VToum43UOS8YxF8WwhXriYGnc1sskbwpXOUDc9uX4+UESzH22P29ovd d8WErY0gPxun8pbJLmxkAtWNhpMvfe0050vk9TL5wqbu9AlbssgTcCXkMQnPw9nC YNN6DDP2lbcBrvgT9YCNL6C+ZKufD52yOQ9qOkwFTEQpjtF4uNtJom+asvlpmS8A vLY9r60wYSvmZhNqBUrj7lyCtXMIu1kkd4w7F77k+DjHoAXyxcUp1DGL51sOmama +TOWWgECgYEA8JtPxP0GRJ+IQkX262jM3dEIkza8ky5moIwUqYdsx0NxHgRRhORT 8c8hAuRBb2G82so8vUHk/fur85OEfc9TncnCY2crpoqsghifKLxrLgtT+qDpfZnx SatLdt8GfQ85yA7hnWWJ2MxF3NaeSDm75Lsm+tBbAiyc9P2jGRNtMSkCgYEAypHd HCctNi/FwjulhttFx/rHYKhLidZDFYeiE/v45bN4yFm8x7R/b0iE7KaszX+Exdvt SghaTdcG0Knyw1bpJVyusavPzpaJMjdJ6tcFhVAbAjm7enCIvGCSx+X3l5SiWg0A R57hJglezIiVjv3aGwHwvlZvtszK6zV6oXFAu0ECgYAbjo46T4hyP5tJi93V5HDi Ttiek7xRVxUl+iU7rWkGAXFpMLFteQEsRr7PJ/lemmEY5eTDAFMLy9FL2m9oQWCg R8VdwSk8r9FGLS+9aKcV5PI/WEKlwgXinB3OhYimtiG2Cg5JCqIZFHxD6MjEGOiu L8ktHMPvodBwNsSBULpG0QKBgBAplTfC1HOnWiMGOU3KPwYWt0O6CdTkmJOmL8Ni blh9elyZ9FsGxsgtRBXRsqXuz7wtsQAgLHxbdLq/ZJQ7YfzOKU4ZxEnabvXnvWkU YOdjHdSOoKvDQNWu6ucyLRAWFuISeXw9a/9p7ftpxm0TSgyvmfLF2MIAEwyzRqaM 77pBAoGAMmjmIJdjp+Ez8duyn3ieo36yrttF5NSsJLAbxFpdlc1gvtGCWW+9Cq0b dxviW8+TFVEBl1O4f7HVm6EpTscdDxU+bCXWkfjuRb7Dy9GOtt9JPsX8MBTakzh3 vBgsyi/sN3RqRBcGU40fOoZyfAMT8s1m/uYv52O6IgeuZ/ujbjY= -----END RSA PRIVATE KEY----- closed返回的是一段ssh私鑰，不難猜想這是下一關連接的私鑰信息，先存起來再說，直接在當前目錄寫發(fā)現(xiàn)沒有權限，這樣我們就需要寫道/tmp目錄下了
bandit16@bandit:~$ mkdir /tmp/ssh_conn bandit16@bandit:~$ vim /tmp/ssh_conn/rsa.priv用這個私鑰去連接第17關。
bandit16@bandit:~$ ssh -i /tmp/ssh_conn/rsa.priv bandit17@localhost Could not create directory '/home/bandit16/.ssh'. The authenticity of host 'localhost (127.0.0.1)' can't be established. ECDSA key fingerprint is SHA256:98UL0ZWr85496EtCRkKlo20X3OPnyPSB5tB5RPbhczc. Are you sure you want to continue connecting (yes/no)? yes Failed to add the host to the list of known hosts (/home/bandit16/.ssh/known_hosts). This is a OverTheWire game server. More information on http://www.overthewire.org/wargames@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @ WARNING: UNPROTECTED PRIVATE KEY FILE! @ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ Permissions 0644 for '/tmp/ssh_conn/rsa.priv' are too open. It is required that your private key files are NOT accessible by others. This private key will be ignored. Load key "/tmp/ssh_conn/rsa.priv": bad permissions bandit17@localhost's password:提示權限太開放了，把權限改600再試。
bandit16@bandit:/tmp/ssh_conn$ ssh -i rsa.priv bandit17@localhost Could not create directory '/home/bandit16/.ssh'. The authenticity of host 'localhost (127.0.0.1)' can't be established. ECDSA key fingerprint is SHA256:98UL0ZWr85496EtCRkKlo20X3OPnyPSB5tB5RPbhczc. Are you sure you want to continue connecting (yes/no)? yes 。 。 。bandit17@bandit:~$ cat /etc/bandit_pass/bandit17 xLYVMN9WE5zQ5vHacb0sZEVqbrp7nBTnLevel 17 → Level 18
Level Goal
There are 2 files in the homedirectory: passwords.old and passwords.new. The password for the next level is in passwords.new and is the only line that has been changed between passwords.old and passwords.new
NOTE: if you have solved this level and see ‘Byebye!’ when trying to log into bandit18, this is related to the next level, bandit19
Commands you may need to solve this level
cat, grep, ls, diff
diff 比較兩個文件的不同, 然后passwd.new不同行行號密碼對應的密碼為bandit18
得到密鑰kfBf3eYk5BPBRzwjqutbbfE887SVc5Yd
Level 18 → Level 19
Level Goal
The password for the next level is stored in a file readme in the homedirectory. Unfortunately, someone has modified .bashrc to log you out when you log in with SSH.
Commands you may need to solve this level
ssh, ls, cat
用上面的密碼，一上來就告訴我byebye,然后自動logout了，搞得我一臉懵逼，題目說是.bashrc文件自動登出的。那我們不分配偽終端就可以了，意思是說禁止分配偽終端。當用ssh或telnet等登錄系統(tǒng)時，系統(tǒng)分配給我們的終端就是偽終端。如果ssh使用此選項登錄系統(tǒng)時，由于禁用，將無法獲得終端；但仍能夠獲得shell，只不過看起來像在本地，也沒有很多應有的環(huán)境變量，例如命令提示符，PS1等。當使用命令ps -ef|grep [b]ash時看到root 22082 22080 0 11:51 ? 00:00:00 -bash顯示終端那里是一個問號。
讀取readme,得到19關的密碼。
Level 19 → Level 20
Level Goal
To gain access to the next level, you should use the setuid binary in the homedirectory. Execute it without arguments to find out how to use it. The password for this level can be found in the usual place (/etc/bandit_pass), after you have used the setuid binary.
Helpful Reading Material
setuid on Wikipedia
先看看家目錄下的文件的權限
屬主的權限為rws, s是特殊權限位，允許一般用戶用root權限執(zhí)行這個文件。
通過文件名是想我們用bandit20這個用戶執(zhí)行這個命令讀取密碼，通過id 命令查看到bandit20用戶的uid為11020，運行這個文件–help 命令查看用法可得用法，最后讀取密碼
Level 20 → Level 21
Level Goal
There is a setuid binary in the homedirectory that does the following: it makes a connection to localhost on the port you specify as a commandline argument. It then reads a line of text from the connection and compares it to the password in the previous level (bandit20). If the password is correct, it will transmit the password for the next level (bandit21).
NOTE: Try connecting to your own network daemon to see if it works as you think
Commands you may need to solve this level
ssh, nc, cat, bash, screen, tmux, Unix ‘job control’ (bg, fg, jobs, &, CTRL-Z, …)
screen命令的說明 https://www.ibm.com/developerworks/cn/linux/l-cn-screen/
tmux是多開終端的命令，job controls經常用就不說了。
這題說是開放一個監(jiān)聽的端口，然后suconnect 文件訪問這個端口如果得到和這關相同的密碼就會返回下一關的密碼，我們就用nc將本關的密碼反饋給連接端口命令如下
成功返回下一關的密碼
Level 21 → Level 22
Level Goal
A program is running automatically at regular intervals from cron, the time-based job scheduler. Look in /etc/cron.d/ for the configuration and see what command is being executed.
Commands you may need to solve this level
cron, crontab, crontab(5) (use “man 5 crontab” to access this)
cron介紹可以參考這篇文章 https://www.cnblogs.com/longjshz/p/5779215.html
先按照提示看看當前目錄下有什么，可以看到這是一個執(zhí)行了一個腳本，然后打開這個腳本看看這是一個定時將22關密碼寫到/tmp目錄下的一個腳本，我們讀取這個臨時文件就知道了下一關的密碼。
Level 22 → Level 23
Level Goal
A program is running automatically at regular intervals from cron, the time-based job scheduler. Look in /etc/cron.d/ for the configuration and see what command is being executed.
NOTE: Looking at shell scripts written by other people is a very useful skill. The script for this level is intentionally made easy to read. If you are having problems understanding what it does, try executing it to see the debug information it prints.
Commands you may need to solve this level
cron, crontab, crontab(5) (use “man 5 crontab” to access this)
先來看看這關所說的定時腳本是什么，如下
實現(xiàn)的功能是取當前用戶名，然后計算 I am user $當前用戶名 的md5值，將bandit22密碼的復制到tmp目錄下的對應的md5值的文件中,讀取
bandit22@bandit:/etc/cron.d$ /bin/bash /usr/bin/cronjob_bandit23.sh Copying passwordfile /etc/bandit_pass/bandit22 to /tmp/8169b67bd894ddbb4412f91573b38db3 bandit22@bandit:/etc/cron.d$ cat /tmp/8169b67bd894ddbb4412f91573b38db3 Yk7owGAcWjwMVRwrTesJEwB7WVOiILLI讀取這個文件，這是本關密碼啊，依次類推，I am user bandit23的hash值就是下一關密碼。
bandit22@bandit:~$ echo I am user bandit23| md5sum 8ca319486bfbbc3663ea0fbe81326349 -bandit22@bandit:~$ cat /tmp/8ca319486bfbbc3663ea0fbe81326349 jc1udXuA1tiHqjIsL8yaapX5XIAI6i0nLevel 23 → Level 24
Level Goal
A program is running automatically at regular intervals from cron, the time-based job scheduler. Look in /etc/cron.d/ for the configuration and see what command is being executed.
NOTE: This level requires you to create your own first shell-script. This is a very big step and you should be proud of yourself when you beat this level!
NOTE 2: Keep in mind that your shell script is removed once executed, so you may want to keep a copy around…
Commands you may need to solve this level
cron, crontab, crontab(5) (use “man 5 crontab” to access this)
老辦法，還是先看看這個定時腳本寫了什么
/var/spool/cron/ 這個目錄下存放的是每個用戶包括root的crontab任務，每個任務以創(chuàng)建者的名字命名，比如tom建的crontab任務對應的文件就是/var/spool/cron/tom。一般一個用戶最多只有一個crontab文件。
我們在/var/spool/bandit24目錄下就可以運行bandit24的定時任務
創(chuàng)建一個放在改目錄下的腳本就可以執(zhí)行了
vim getpass.sh進入vim編輯模式輸入腳本
寫入
cat /etc/bandit_pass/bandit24 > /tmp/bandit24pass
：wq保存退出
這時候在/var/spool/bandit24目錄下不一定能看見你寫的腳本，就像前面的定時任務腳本里面寫的，執(zhí)行完腳本這個就任務就刪除了，所以沒看到也不要奇怪。
這個時候說明我們的腳本已經執(zhí)行了，可以去/tmp目錄查看我們的密碼了
下一關的密鑰已經寫好了
Level 24 → Level 25
Level Goal
A daemon is listening on port 30002 and will give you the password for bandit25 if given the password for bandit24 and a secret numeric 4-digit pincode. There is no way to retrieve the pincode except by going through all of the 10000 combinations, called brute-forcing.
根據python的pwntools寫個腳本跑密碼就好了，注意，在其他目錄下我們是沒有寫權限的，這個腳本只能在/tmp目錄下創(chuàng)建。如果用的是我下面這種receive line方法，有些破壞輸出的結果我要多接收一行過濾掉，
vim /tmp/conn.py
創(chuàng)建腳本如下：
終端運行python /tmp/conn.py
[+] Trying pincode: 0377 Wrong! Please enter the correct pincode. Try again.[+] Trying pincode: 0378 Correct!Got Pincode: 0378 The password of user bandit25 is uNG9O58gUE7snukf3bvZ0rxhtnjzSGzG輸出結果如上所示，前面其他猜解過程就不貼上來了。
Level 25 → Level 26
Level Goal
Logging in to bandit26 from bandit25 should be fairly easy… The shell for user bandit26 is not /bin/bash, but something else. Find out what it is, how it works and how to break out of it.
Commands you may need to solve this level
ssh, cat, more, vi, ls, id, pwd
登錄上去可以看到家目錄上面有一個bandit26.sshkey, 可以像之前一樣用這個私鑰文件去連接遠程的主機, ssh -i bandit26.sshkey bandit26@localhost, 發(fā)現(xiàn)連接直接被遠程關閉了，加上-T 參數(shù)也沒有用，題目也提示說這個用的是其他shell, 查看其某用戶用的什么shell 可以查看/etc/passwd。
passwd文件的格式為:
賬號名稱：即登陸時的用戶名
密碼：早期UNIX系統(tǒng)的密碼是放在這個文件中的，但因為這個文件的特性是所有程序都能夠讀取，所以，這樣很容易造成數(shù)據被竊取，因此后來就將這個字段的密碼數(shù)據改放到/etc/shadow中了
UID：用戶ID，每個賬號名稱對應一個UID，通常UID=0表示root管理員
GID：組ID，與/etc/group有關，/etc/group與/etc/passwd差不多，是用來規(guī)范用戶組信息的
用戶信息說明欄： 用來解釋這個賬號是干什么的
家目錄：home目錄，即用戶登陸以后跳轉到的目錄，以root用戶為例，/root是它的家目錄，所以root用戶登陸以后就跳轉到/root目錄這里
Shell：用戶使用的shell，通常使用/bin/bash這個shell，這也就是為什么登陸Linux時默認的shell是bash的原因，就是在這里設置的，如果要想更改登陸后使用的shell，可以在這里修改。另外一個很重要的東西是有一個shell可以用來替代讓賬號無法登陸的命令，那就是/sbin/nologin。
那bandit26用戶用到的shell就是/usr/bin/showtext
bandit25@bandit:~$ cat /usr/bin/showtext #!/bin/shexport TERM=linuxmore ~/text.txt exit 0系統(tǒng)關閉連接的原因是這個exit 0, 在這個exit 之前執(zhí)行我們想要的命令就可以達到我們想要的效果了。
在more 命令執(zhí)行之前可以執(zhí)行命令即可，把會話的終端縮小，然后用文件連接bandit26，這樣可以出發(fā)自動more, 在more命令還沒有結束的時候按v進入vim編輯模式。再就是用vim特有的:e file，vim模式下的e命令可以導入文件到編輯器內，我們知道密碼的所在，因此就可以用e命令來導入密碼文件
然后26關的密鑰就被導入到終端可讀取了，密鑰為
5czgV9L3Xx8JPOyRbXh6lQbmIOWvPT6Z
Level 26 → Level 27
Level Goal
Good job getting a shell! Now hurry and grab the password for bandit27!
Commands you may need to solve this level
ls
這一關使用密碼ssh登陸之后也是直接斷開了，所以跟上一關套路一樣，進入more模式，利用vim模式執(zhí)行命令，這次不能用e來讀取文件了，因為權限不夠。!command也不行，!sh也不行，后來查看資料發(fā)現(xiàn)vim還有一種需要先設置shell的目錄才行
vim模式下
然后設置完成上去就可以登錄了。ls一下
bandit26@bandit:~$ ls bandit27-do text.txt有個bandit27-do文件，執(zhí)行這個文件讀取bandit27就可以了。
bandit26@bandit:~$ ./bandit27-do cat /etc/bandit_pass/bandit27 3ba3118a22e93127a4ed485be72ef5eaLevel 27 → Level 28
Level Goal
There is a git repository at ssh://bandit27-git@localhost/home/bandit27-git/repo. The password for the user bandit27-git is the same as for the user bandit27.
Clone the repository and find the password for the next level.
Commands you may need to solve this level
git
這題是主要是克隆項目的命令，直接在當前目錄是新建不了新文件的，所以我們在臨時目錄下創(chuàng)建目錄即可，具體步驟如下，發(fā)現(xiàn)這個項目的里面的README就是存儲的的密鑰
Level 28 → Level 29
Level Goal
There is a git repository at ssh://bandit28-git@localhost/home/bandit28-git/repo. The password for the user bandit28-git is the same as for the user bandit28.
Clone the repository and find the password for the next level.
Commands you may need to solve this level
git
克隆項目的過程和之前一樣
題目告訴我們這次的密碼是寫在某個文件里面了,git log查看提交歷史，然后對應版本提交id, 查找區(qū)別，得出密碼。
bandit28@bandit:/tmp/conn28/repo$ git log commit 073c27c130e6ee407e12faad1dd3848a110c4f95 Author: Morla Porla <morla@overthewire.org> Date: Tue Oct 16 14:00:39 2018 +0200fix info leakcommit 186a1038cc54d1358d42d468cdc8e3cc28a93fcb Author: Morla Porla <morla@overthewire.org> Date: Tue Oct 16 14:00:39 2018 +0200add missing datacommit b67405defc6ef44210c53345fc953e6a21338cc7 Author: Ben Dover <noone@overthewire.org> Date: Tue Oct 16 14:00:39 2018 +0200initial commit of README.md bandit28@bandit:/tmp/conn28/repo$ git diff 186a 073c diff --git a/README.md b/README.md index 3f7cee8..5c6457b 100644 --- a/README.md +++ b/README.md @@ -4,5 +4,5 @@ Some notes for level29 of bandit.## credentials- username: bandit29 -- password: bbc96594b4e001778eee9975372716b2 +- password: xxxxxxxxxxLevel 29 → Level 30
Level Goal
There is a git repository at ssh://bandit29-git@localhost/home/bandit29-git/repo. The password for the user bandit29-git is the same as for the user bandit29.
Clone the repository and find the password for the next level.
Commands you may need to solve this level
git
git show命令，git log命令還有git diff命令查看git 提交歷史，利用git branch -a 命令可以查詢分支，發(fā)現(xiàn)總共有四個分支。
git checkout 可以切換分支，當切換到dev查看gitlog 可以發(fā)現(xiàn)，最新的版本里面有個data needed for development
bandit29@bandit:/tmp/conn29/repo$ git checkout dev Switched to branch 'dev' Your branch is up-to-date with 'origin/dev'. bandit29@bandit:/tmp/conn29/repo$ git log commit 33ce2e95d9c5d6fb0a40e5ee9a2926903646b4e3 Author: Morla Porla <morla@overthewire.org> Date: Tue Oct 16 14:00:41 2018 +0200add data needed for developmentcommit a8af722fccd4206fc3780bd3ede35b2c03886d9b Author: Ben Dover <noone@overthewire.org> Date: Tue Oct 16 14:00:41 2018 +0200add gif2asciicommit 84abedc104bbc0c65cb9eb74eb1d3057753e70f8 Author: Ben Dover <noone@overthewire.org> Date: Tue Oct 16 14:00:41 2018 +0200fix usernamecommit 9b19e7d8c1aadf4edcc5b15ba8107329ad6c5650 Author: Ben Dover <noone@overthewire.org> Date: Tue Oct 16 14:00:41 2018 +0200initial commit of README.md然后在這個版本里面的README發(fā)現(xiàn)密碼
bandit29@bandit:/tmp/conn29/repo$ cat README.md # Bandit Notes Some notes for bandit30 of bandit.## credentials- username: bandit30 - password: 5b90576bedb2cc04c86a9e924ce42fafLevel 30 → Level 31
Level Goal
There is a git repository at ssh://bandit30-git@localhost/home/bandit30-git/repo. The password for the user bandit30-git is the same as for the user bandit30.
Clone the repository and find the password for the next level.
Commands you may need to solve this level
git
git show-ref可以現(xiàn)實本地存儲庫的所有可用的引用以及關聯(lián)的提交ID
Level 31 → Level 32
Level Goal
There is a git repository at ssh://bandit31-git@localhost/home/bandit31-git/repo. The password for the user bandit31-git is the same as for the user bandit31.
Clone the repository and find the password for the next level.
Commands you may need to solve this level
git
這題是讓我們提交到遠程倉庫
得到下一關的密鑰56a9bf19c63d650ce78e6ec0354ee45e
Level 32 → Level 33
After all this git stuff its time for another escape. Good luck!
Commands you may need to solve this level
sh, man
連接的最后直接給了你一個大寫的終端。怎么辦呢，我還沒思路
--------------------2019.5.13 更新-----------------
感謝評論區(qū)小伙伴提醒$0可以進入正常終端
總結
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