Python 3.6.5的坑 AES padding

Posted 2021-01-27 luoyueren

from Crypto.Cipher import AES

key = ‘DF11-FB15-B7B2-15AB-47B7-7AC4-C6F9-5EFE‘
cryptor = AES.new(key.encode(‘utf-8‘),AES.MODE_CBC,str(key[0:16]).encode(‘utf-8‘))

text = b‘1234567890abc‘
encrypted= cryptor.encrypt(pad_text(text));

 

 

def pad_text(s):

‘‘‘Pad an input string according to PKCS#7‘‘‘
BS = AES.block_size
return s + (BS - len(s) % BS) * chr(BS - len(s) % BS).encode("utf-8")

"""
注意啦，
1）当text是二进制流时， chr(BS - len(s) % BS).encode("utf-8")里的".encode("utf-8")"是不能少的， 不然会有如下错误：
TypeError: can‘t concat str to bytes
2）Key必须要.encode(‘utf-8‘)： key.encode(‘utf-8‘) ， 32位的key
3）IV必须要.encode(‘utf-8‘) ： str(key[0:16]).encode(‘utf-8‘)，从32位的key取16位当做IV的输入值

 

#Python 3.6.5 (v3.6.5:f59c0932b4, Mar 28 2018, 17:00:18) [MSC v.1900 64 bit (AMD64)] on win32
#Crypto 1.4.1
"""