Question

pycrypto AES CBC加密中的未知错误，密钥被剥离为16字节，但仍然会出现此错误：

来源帮助文章： http://support.ideascale.com/kb/ideascale-setup/single-sign-on-multipass-token-based 更新类模块： class multipassGenerator（）： def init （个体经营）： api_key =“61fd537c-8961-479f-a413-aa91a401c3f5-5e866a4c-09c7-4ba0-8da1-cd8e4e88e3ec” app_key =“25151”

    def generateMultipass(self,usrEmail,usrName):
        if usrEmail and usrName:
            # EXAMPLE: Multipass JSON Token
            message = {"email":usrEmail,"name":usrName,"expires":"2099-02-24T19:55:31.111-08:00"}
            block_size = 16
            mode = AES.MODE_CBC

            # Replace api_key and app_key with your Community's API Key and Site Key
            # The keys below are for this community: http://multipass.ideascale.com/
            # Do not share your keys - the keys below are for testing purposes only.
            # See: http://support.ideascale.com/kb/ideascale-setup/single-sign-on-multipass-token-based
            api_key = self.api_key
            app_key = self.app_key

            json = JSON.dumps(message, separators=(',',':'))

            salted = api_key+app_key
            saltedHash = hashlib.sha1(salted).digest()[:16]
            logging.info(message)
            logging.info(salted)
            #logging.info(saltedHash)
            logging.info(len(saltedHash))
            json_bytes = array.array('b', json[0 : len(json)]) 

            pad = block_size - len(json_bytes.tostring()) % block_size
            data = json_bytes.tostring() + pad * chr(pad)
            aes = AES.new(saltedHash, mode)
            encrypted_bytes = aes.encrypt(data)

            b64token = base64.b64encode(encrypted_bytes)
            b64token = re.sub(r'\s+' ,'' ,b64token)
            b64token = re.sub(r'\=+$','' ,b64token)
            b64token = re.sub(r'\+'  ,'-',b64token)
            b64token = re.sub(r'\/'  ,'_',b64token)
            token = urllib.quote(b64token)

            return token

我们执行此操作：

from getUserMuiltipass import multipassGenerator
tokenGenerator = multipassGenerator()
tokenGenerator.api_key = u"61fd537c-8961-479f-a413-aa91a401c3f5-5e866a4c-09c7-4ba0-8da1-cd8e4e88e3ec"
tokenGenerator.app_key = u"25151"
email = 'sanvvvdeepkffffs@advaiddddya.com' # any input same issue 
print  email.split('@')[0]
ssoToken = tokenGenerator.generateMultipass(email,email.split('@')[0]) 
print ssoToken

此部分代码导致问题：AES.new（saltedHash，mode）

Traceback (most recent call last):
  File "C:\Program Files (x86)\Google\google_appengine\google\appengine\tools\devappserver2\python\request_handler.py", line 156, in handle_interactive_request
    exec(compiled_code, self._command_globals)
  File "<string>", line 7, in <module>
  File "C:\appengine\ideasONSMS\getUserMuiltipass.py", line 59, in generateMultipass
    aes = AES.new(saltedHash, mode)
  File "C:\Python27\lib\site-packages\Crypto\Cipher\AES.py", line 95, in new
    return AESCipher(key, *args, **kwargs)
  File "C:\Python27\lib\site-packages\Crypto\Cipher\AES.py", line 59, in __init__
    blockalgo.BlockAlgo.__init__(self, _AES, key, *args, **kwargs)
  File "C:\Python27\lib\site-packages\Crypto\Cipher\blockalgo.py", line 141, in __init__
    self._cipher = factory.new(key, *args, **kwargs)
ValueError: IV must be 16 bytes long

Answer 1

可能是因为AES.new fn对于CBC模式需要3个参数：

AES.new(key, *args, **kwargs)

3-d arg是IV。可以像这样生成：

iv = Random.new().read(AES.block_size)

AES CBC加密pycrypto 2.6中的未知错误

1 个答案: