我在PHP中有一个函数加密文本,如下所示:
function encrypt($text)
{
$Key = "MyKey";
return trim(base64_encode(mcrypt_encrypt(MCRYPT_RIJNDAEL_256, $Key, $text, MCRYPT_MODE_ECB, mcrypt_create_iv(mcrypt_get_iv_size(MCRYPT_RIJNDAEL_256, MCRYPT_MODE_ECB), MCRYPT_RAND))));
}
如何在Python中解密这些值?
答案 0 :(得分:16)
要解密这种加密形式,您需要获得Rijndael的版本。可以找到一个here。然后,您将需要模拟PHP Mcrypt模块中使用的键和文本填充。他们添加'\0'
来填充文本并键入正确的大小。它们使用的是256位块大小,并且您使用的密钥使用的密钥大小为128(如果您给它一个更大的密钥,它可能会增加)。不幸的是,我链接的Python实现一次只编码一个块。我创建了python函数,用于模拟Python中的加密(用于测试)和解密
import rijndael
import base64
KEY_SIZE = 16
BLOCK_SIZE = 32
def encrypt(key, plaintext):
padded_key = key.ljust(KEY_SIZE, '\0')
padded_text = plaintext + (BLOCK_SIZE - len(plaintext) % BLOCK_SIZE) * '\0'
# could also be one of
#if len(plaintext) % BLOCK_SIZE != 0:
# padded_text = plaintext.ljust((len(plaintext) / BLOCK_SIZE) + 1 * BLOCKSIZE), '\0')
# -OR-
#padded_text = plaintext.ljust((len(plaintext) + (BLOCK_SIZE - len(plaintext) % BLOCK_SIZE)), '\0')
r = rijndael.rijndael(padded_key, BLOCK_SIZE)
ciphertext = ''
for start in range(0, len(padded_text), BLOCK_SIZE):
ciphertext += r.encrypt(padded_text[start:start+BLOCK_SIZE])
encoded = base64.b64encode(ciphertext)
return encoded
def decrypt(key, encoded):
padded_key = key.ljust(KEY_SIZE, '\0')
ciphertext = base64.b64decode(encoded)
r = rijndael.rijndael(padded_key, BLOCK_SIZE)
padded_text = ''
for start in range(0, len(ciphertext), BLOCK_SIZE):
padded_text += r.decrypt(ciphertext[start:start+BLOCK_SIZE])
plaintext = padded_text.split('\x00', 1)[0]
return plaintext
可以使用如下:
key = 'MyKey'
text = 'test'
encoded = encrypt(key, text)
print repr(encoded)
# prints 'I+KlvwIK2e690lPLDQMMUf5kfZmdZRIexYJp1SLWRJY='
decoded = decrypt(key, encoded)
print repr(decoded)
# prints 'test'
为了比较,这里是PHP的输出,文本相同:
$ php -a
Interactive shell
php > $key = 'MyKey';
php > $text = 'test';
php > $output = mcrypt_encrypt(MCRYPT_RIJNDAEL_256, $key, $text, MCRYPT_MODE_ECB);
php > $encoded = base64_encode($output);
php > echo $encoded;
I+KlvwIK2e690lPLDQMMUf5kfZmdZRIexYJp1SLWRJY=
答案 1 :(得分:4)
如果您愿意在PHP端使用MCRYPT_RIJNDAEL_128而不是256,那么这很简单:
from Crypto.Cipher import AES
import base64
key="MyKey"
def decrypt(text)
cipher=AES.new(key)
return cipher.decrypt(base64.b64decode(text))
答案 2 :(得分:2)
虽然@ 101100的答案在当时是一个很好的答案,但它已不再可行。该引用现在是一个断开的链接,代码只能在较旧的Pythons(< 3)上运行。
相反,pprp project似乎很好地填补了空白。在Python 2或Python 3上,只需pip install pprp
,然后:
import pprp
import base64
KEY_SIZE = 16
BLOCK_SIZE = 32
def encrypt(key, plaintext):
key = key.encode('ascii')
plaintext = plaintext.encode('utf-8')
padded_key = key.ljust(KEY_SIZE, b'\0')
sg = pprp.data_source_gen(plaintext, block_size=BLOCK_SIZE)
eg = pprp.rjindael_encrypt_gen(padded_key, sg, block_size=BLOCK_SIZE)
ciphertext = pprp.encrypt_sink(eg)
encoded = base64.b64encode(ciphertext)
return encoded.decode('ascii')
def decrypt(key, encoded):
key = key.encode('ascii')
padded_key = key.ljust(KEY_SIZE, b'\0')
ciphertext = base64.b64decode(encoded.encode('ascii'))
sg = pprp.data_source_gen(ciphertext, block_size=BLOCK_SIZE)
dg = pprp.rjindael_decrypt_gen(padded_key, sg, block_size=BLOCK_SIZE)
return pprp.decrypt_sink(dg).decode('utf-8')
key = 'MyKey'
text = 'test'
encoded = encrypt(key, text)
print(repr(encoded))
# prints 'ju0pt5Y63Vj4qiViL4VL83Wjgirq4QsGDkj+tDcNcrw='
decoded = decrypt(key, encoded)
print(repr(decoded))
# prints 'test'
我有点沮丧的是,密文与你在101100的回答中看到的不同。但是,我已经使用这种技术成功解密了在PHP中加密的数据,如OP中所述。