1
,我现在有将列出所有可能的组合蛮力黑客概念在Python 3
from itertools import product
password = "hr"
chars = 'abcdefghijklmnopqrstuvwxyz' #characters to look for, can have letters added
for length in range(1, 3): #only do lengths of 1 + 2 - namely from aa - zz not onto aaa
to_attempt = product(chars, repeat=length)
for attempt in to_attempt:
print(''.join(attempt))
我需要做的就是把每尝试尝试,它与变量“密码”比较的代码,如果它匹配突围for循环其他进行,任何想法?
你的第5行不正确,它应该读取'if''.join(attempt)== password:'按照原始代码。 – phantom 2014-11-01 20:44:10
@phantom你能稍微解释一下吗?对于任何字符串,==''.join(a)。你为什么要这样做,而不是尝试==密码? – furkle 2014-11-01 20:46:17
因为变量'attempt'不是两个字符的密码,所以它是单个字符。 – phantom 2014-11-01 20:52:27