如何用密钥和iv解密AES加密的数据

Question

我正在尝试解密AES加密的数据。 我带着钥匙和静脉注射，但我想钥匙也是编码的。在

以下是我使用的代码：

import binascii
from Crypto.Cipher import AES
enckey = '5f35604280b44dd1073f7ee83e346d81'
key = binascii.unhexlify(enckey)
key32 = "{: <32}".format(key).encode("utf-8")
data='692fa1deafad8ad80b98cd6f077899e9be457ac5364c3822aae9457d4912e4829d71cb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'
cipher_text = binascii.unhexlify(data)
# Decryption
decryption_suite = AES.new('key32', AES.MODE_CBC, 'heF9BATUfWuISyO8')
plain_text = decryption_suite.decrypt(cipher_text)
print plain_text

这是我收到的错误：

^{pr2}$

我无法解决我的钥匙和数据的问题。。。 如果我需要对代码进行任何更改，请告诉我。在

Answer 1

以下代码有效：

from Crypto.Cipher import AES

keyAscii = '5f35604280b44dd1073f7ee83e346d81'
keyBinary = bytes(keyAscii, 'ascii')
ciphertextHex='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'
ciphertextBinary = bytes.fromhex(ciphertextHex)
ivAscii = 'heF9BATUfWuISyO8'
ivBinary = bytes(ivAscii, 'ascii')
# Decryption
decrypter = AES.new(keyBinary, AES.MODE_CBC, ivBinary)
plaintextBinary = decrypter.decrypt(ciphertextBinary)
plaintext = plaintextBinary.decode('utf-8')
print(plaintext)

和输出

^{pr2}$

---

但它在很多方面都是错误的：

• 密钥看起来像32个字符的十六进制字符串（它将编码128位的密钥），但实际上您需要将其视为ASCII编码的32字节密钥。这是错误的，因为密钥应该是均匀分布的二进制字符串
• 类似地，IV应该是一个统一的二进制文件，但实际上它是ASCII
• IV是固定的，但是对于每个消息而言，IV的整个点是不同的（对于CBC模式来说是不可预测的随机性）。在
• CBC模式易受padding oracle攻击的攻击，如果主动攻击者能够执行选定的密文攻击，并在其中得知哪个密文成功解密，则允许主动攻击者恢复明文。在

您应该对唯一的IVs使用经过身份验证的加密。在

Answer 2

这个可能对你有帮助

import binascii
from Crypto.Cipher import AES
import re

enckey = '5f35604280b44dd1073f7ee83e346d81'
key32 = "{: <32}".format(enckey).encode("utf-8")
cipher = AES.new(key32, AES.MODE_ECB)
data='692fa1deafad8ad80b98cd6f077899e9be457ac5364c3822aae9457d4912e4829d71cb8702bd10e1d54f7a0461edba193517b353835480bd174804f586776e623473022548ff098a9545b608282bf498a36968dd6b858ad631f6eaa79ea1a87c984f4a8da5a9d1cee1b11b32d46c0d2a670d4e634ecc47c7105387a0a38853c91e10747170de69ebf6f0e1a99f0134ddb0af0cec2cfc70f53c9eab7227460cf1153ef686a5dc5014bd286fb0efdec571327f5a4874bec5fd5c65f09f0ed10e906e4199dd8c3cb8340aca1904f486a70b02554581f0e723d22854188e933ed9fce60172099bc675b89eba39651bbc0658ae264213217f14ff4f0824494585d8856dfd44e4ce9505e43762a9f1ea48f9c736603e83c3e10c5740cdf279dc3a914e19eee089160ffa91180d1b429938ab1b6a4272d1779f7702f760cbac3f35fc35c16fcf21c7e00183f306e7a18f71ffb3b62b91250dca7dd627876a6cedbfe83f0f18abbbb7c7650566a7f761844243fe1271cef22b1026a3f1d37b8e7bd7c068331897680ec101e269ce66c3f129de33d3277c2cc120feb88f77f1bb851d41b83468128366b7ed92ae07f37675cee07355ebcdfcba90a690e3d4817cd18123a0c9de175ea6c5049c51170ee73facd5148f6525024116991b0601598a3501e770493dab0653e146981e91d2ea9c50fbd1e6b8bb38407655c518f30852ce43ed62d1c578e642c4fa92f00bbf102c3418ed52ed23138c86d327bbc4718ec44440f289e3af6c096c7ad69af5d941768b0f4b2e3decfad5dcfe6dc491ce4f2f9d86d226b87f19dfb56dc44f6d66820773e6fcfa4fcd7958da2d63903762705799a414baf93081242c2b594981c93b892f4f28883203875a4010ace9a5eafea51ee406'
cipher_text = binascii.unhexlify(data)
# Decryption


plain_text = re.sub('\0*$','', cipher.decrypt( data[16:]))
print plain_text

或者试试这个

^{pr2}$