加密解密
在线网站
| 网址 | 解释 |
|---|---|
|
https://gchq.github.io/CyberChef/ |
CyberChef（网络厨师） |
|
https://www.dcode.fr/ |
加密解密、数学、代码、在线计算 |
|
https://crackstation.net/ |
免费密码HASH破解器 |
|
https://www.base64encode.org/ |
Base64编码 |
|
https://www.base64decode.org/ |
Base64解码 |
|
https://rumkin.com/tools/cipher/ |
Rumkin密码和代码解密工具 |
|
https://www.unphp.net |
在线PHP代码解密 |
|
https://cmd5.com/ |
MD5在线解密破解 |
|
https://www.somd5.com/ |
MD5免费在线解密破解 |
|
https://www.boxentriq.com/ |
密码破译和逻辑谜题解决工具 |
|
https://quipqiup.com/ |
自动化密码破解 |
|
https://www.nayuki.io/page/automatic-caesar-cipher-breaker-javascript |
自动凯撒密码破解 |
Fernet解密
https://asecuritysite.com/encryption/ferdecodefrom cryptography.fernet import Fernet
key = ""
token = ""
cipher = Fernet(key)
decoded = cipher.decrypt(token)Malbloge解密
http://www.malbolge.doleczek.pl/
https://zb3.me/malbolge-tools/Dvorak Keyboard解密
https://www.geocachingtoolbox.com/index.php?lang=en&page=dvorakKeyboardDTFM解密
http://dl.djsoft.net/DTMFChecker.zip
https://www.dcode.fr/prime-numbers-cipherBcrypt解密
git clone https://github.com/BREAKTEAM/Debcrypt
python3 crack.py
Cistercian Monk解密
https://www.dcode.fr/cistercian-numbersVigenere解密
https://www.dcode.fr/vigenere-cipher多次点击电话代码转换
https://www.dcode.fr/code-multitap-abc
http://rumkin.com/tools/cipher/atbash.php异或消息解密
git clone https://github.com/laconicwolf/crypto-tools
python3 crack_repeating_key_xor.py -f <file> -xRSA PKCS#1解密
from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_v1_5
import cryptastib as ca
ca.queries = 0
# key = RSA.generate(1024, e=3)
# print repr((key.n, key.e, key.d))
(N, e, d) = (151242689083100816738181002620937999512181809379387195583593491291603542900108691523222285405372857139696327669320405999262960106462050217947074001784527818795198802600940780900545580866814212880284349556332186821342122597851846151778484011876082986461493379210023160659498744772656523909821057886586431311599L, 3L, 100828459388733877825454001747291999674787872919591463722395660861069028600072461015481523603581904759797551779546937332841973404308033478631382667856351862771810527368138135852999223344751183054387918163104751737882143190101541038691020249177008890480468119734661689468498909557170851608519691070322297440427L)
key = RSA.construct((N, e, d))
plaintext = '\x02BJ\xe2s^\x1a\x9f\xcfAS\xceGZ\xa0\x99n\xae\x1d;\xd6N\xbb\xf6\xcd\xaf\xb5\'$\x81/0w\xf8\x88"\xb6\xda\xbb\x86\xb6\x9b\xb4z\xde\x04\xd4\xbd\xcf\x17\xe5\xe3G#\x1fv\xabP\x17$015\xb5/il\x12U\xdf0\x1b\xdcEl\x0fQw`\x02#\xd4kQ\x1a/\x89\xfa\x15\x04U\xef>\x90v\xee\x01O\xde\x9d\x0bi\x17\xd1\x16\xe2\x8b\xfa\x087\xb3\x83\x00test plaintext'
ciphertext = key.encrypt(plaintext, 0)[0]
def oracle(ciphertext):
plaintext = key.decrypt(ciphertext)
ca.queries += 1
return plaintext.encode('hex')[:2] == '02'
print 'Testing Bleichenbacher\'s oracle...'
decrypted = ca.bb98_padding_oracle(ciphertext, oracle, key.e, key.n, verbose=True, debug=False)
print "Plaintext is %r" % key.decrypt(ciphertext)
print "Attack produced plaintext of %r" % (decrypted)
if decrypted != key.decrypt(ciphertext):
print "Failed with %d queries" % ca.queries
raise BleichenbacherAttackFailedError
else:
print "Succeeded with %d queries" % ca.queries无需私钥，使用公钥解密enc文件
git clone https://github.com/RsaCtfTool/RsaCtfTool
python3 RsaCtfTool.py --publickey ./key.pub --private
openssl rsautl -decrypt -inkey key.pri -in flag.enc -out flag.txt终端中解密Base64
echo 'R2hvc3RfV29sZl9MYWI=' | base64 -d