如何使用已提供的密鑰和iv解密AES加密數據

Question

我試圖解密AES加密數據。 我有鑰匙和iv我，但我認爲鑰匙也編碼。

這裏是我使用的代碼：

import binascii 
from Crypto.Cipher import AES 
enckey = '5f35604280b44dd1073f7ee83e346d81' 
key = binascii.unhexlify(enckey) 
key32 = "{: <32}".format(key).encode("utf-8") 
data='692fa1deafad8ad80b98cd6f077899e9be457ac5364c3822aae9457d4912e4829d71cb 
8702bd10e1d54f7a0461edba193517b353835480bd174804f586776e623473022548ff098a95 
45b608282bf498a36968dd6b858ad631f6eaa79ea1a87c984f4a8da5a9d1cee1b11b32d46c0d 
2a670d4e634ecc47c7105387a0a38853c91e10747170de69ebf6f0e1a99f0134ddb0af0cec2c 
fc70f53c9eab7227460cf1153ef686a5dc5014bd286fb0efdec571327f5a4874bec5fd5c65f0 
9f0ed10e906e4199dd8c3cb8340aca1904f486a70b02554581f0e723d22854188e933ed9fce6 
0172099bc675b89eba39651bbc0658ae264213217f14ff4f0824494585d8856dfd44e4ce9505 
e43762a9f1ea48f9c736603e83c3e10c5740cdf279dc3a914e19eee089160ffa91180d1b4299 
38ab1b6a4272d1779f7702f760cbac3f35fc35c16fcf21c7e00183f306e7a18f71ffb3b62b91 
250dca7dd627876a6cedbfe83f0f18abbbb7c7650566a7f761844243fe1271cef22b1026a3f1 
d37b8e7bd7c068331897680ec101e269ce66c3f129de33d3277c2cc120feb88f77f1bb851d41 
b83468128366b7ed92ae07f37675cee07355ebcdfcba90a690e3d4817cd18123a0c9de175ea6 
c5049c51170ee73facd5148f6525024116991b0601598a3501e770493dab0653e146981e91d2 
ea9c50fbd1e6b8bb38407655c518f30852ce43ed62d1c578e642c4fa92f00bbf102c3418ed52 
ed23138c86d327bbc4718ec44440f289e3af6c096c7ad69af5d941768b0f4b2e3decfad5dcfe 
6dc491ce4f2f9d86d226b87f19dfb56dc44f6d66820773e6fcfa4fcd7958da2d639037627057 
99a414baf93081242c2b594981c93b892f4f28883203875a4010ace9a5eafea51ee406' 
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 

這是我收到的錯誤：

UnicodeDecodeError: 'ascii' codec can't decode byte 0x80 in position 4: 
ordinal not in range(128) 

我無法用我的鑰匙和數據找出問題.. 如果我必須對我的代碼進行任何更改，請讓我知道。

Answer 1

這一個可以幫助你

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='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' 
cipher_text = binascii.unhexlify(data) 
# Decryption 


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

或試試這個

import binascii 
from Crypto.Cipher import AES 


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 = cipher.decrypt(cipher_text) 
print plain_text 

Answer 2

下面的代碼工作：

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) 

和輸出

connection_type=wifi&android_id=863e87fea9a09533&app_name=AstroNest&app_version=53&app_version_name=2.7.1&device_brand=motorola&device_cpu_type=armv7l&device_model=XT1562&google_aid=ab95a01a-242b-4ac2-ad12-b6189e983a56&google_ad_tracking_disabled=0&insdate=1494826343&installer=com.android.vending&language=en&mac_address=02%3A00%3A00%3A00%3A00%3A00&mat_id=45c0a743-a948-434b-a20d-fe66e870d285&os_version=6.0.1&screen_density=3.0&screen_layout_size=1920x1080&sdk_version=3.11.4&conversion_user_agent=Dalvik%2F2.1.0+%28Linux%3B+U%3B+Android+6.0.1%3B+XT1562+Build%2FMPDS24.107-70-1-5%29&currency_code=USD&revenue=0.0&system_date=1494826532 

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但它在許多方面加密錯誤：

• 的關鍵看起來像32字符的十六進制字符串（這將編碼128位密鑰），但實際上你需要把它當作ASCII編碼32字節鍵。這是錯誤的，因爲密鑰應該是均勻分佈的二進制字符串
• 類似地，IV應該是一個統一的二進制文件，但它實際上是ASCII
• IV是固定的，但IV的整個點是不同的（對於CBC模式是不可預測的隨機）每條消息。
• CBC模式易受填充Oracle攻擊的影響，如果攻擊者可以執行選擇的密文攻擊，並知道哪個密文成功解密，攻擊者可以恢復明文。

您應該使用帶有唯一IV的身份驗證加密代替。