这是我的第一个问题，所以希望我这样做。我正在开展一个猜测＆＃39;的项目。用户输入的密码。该程序已经知道密码，但我试图让它根据密码的类型循环遍历所有可能的字符组合。该程序还会查找密码是数字，字母数字还是字母数字并包含特殊字符。
到目前为止我还没有遇到太多麻烦（它可以成功找到一个数字密码）但是在尝试了几个小时来配置一个字母数字猜测功能后，我决定寻求一些帮助
我的问题是：我该如何实施？我已经附加了我的程序，该程序分为两个文件。 main.py是您应该运行的文件。 Guesser.py中的第43到75行是问题区域。我对Python比较陌生，这是我做过的规模最大的项目，所以我很欣赏对设计的任何建议或改变！ （如果有什么事情没有意义，请告诉我）
MAIN.PY
import string
import Guesser
letters = list(string.printable)
del(letters[94:100])
class Main():
def __init__():
qst = input('One password (1) or multiple (m)? ')
if qst == '1':
pss = input('Password: ')
if len(pss) < 4:
print('Error! Password too short!')
Main.__init__()
elif len(pss) > 15:
print('Error! Password too long!')
Main.__init__()
else:
pass
Main.type_check(qst, pss)
else:
e = 0
print('Type \'done\' if you have typed in all of your passwords')
plist = list()
c = True
while c == True:
e = e + 1
i = input('Password %s: ' % e)
i.replace('\n', '')
if len(i) < 4:
print('Error! %s is too short.' % i)
Main.__init__()
elif len(i) > 15:
print('Error! %s is too long.' % i)
Main.__init__()
if i != 'done':
plist.append(i)
else:
c = False
pss = plist
Main.type_check(qst, pss)
def type_check(qst, pss):
if qst == '1':
try:
int(pss)
pass_type = 'num'
Guesser.Guesser.check(pss, pass_type)
except ValueError:
if any(every in pss for every in list(string.printable)) == False:
print('Password contains invalid characters')
Main.__init__()
if any(punct in pss for punct in list(string.punctuation)):
pass_type = 'alphasp'
Guesser.Guesser.check(pss, pass_type)
else:
pass_type = 'alpha'
Guesser.Guesser.check(pss, pass_type)
else:
pass_dict = dict()
for x in range(0, len(pss)):
try:
int(pss[x])
pass_dict[pss[x]] = 'num'
except ValueError:
if any(every in pss[x] for every in list(string.printable)) == False:
print('Password contains invalid characters')
Main.__init__()
if any(punct in pss[x] for punct in list(string.punctuation)):
pass_dict[pss[x]] = 'alphasp'
else:
pass_dict[pss[x]] = 'alpha'
pss = pass_dict
pass_type = None
Guesser.Guesser.check(pss, pass_type)
if __name__ == '__main__':
print('Your password(s) has to be 4 digits or longer and 15 digits or shorter; no spaces')
Main.__init__()
GUESSER.PY
import main
import time
import string
class Guesser():
#fixed
def check(pss, pass_type):
if pass_type != None:
if pass_type == 'num':
guess_param = 'n'
elif pass_type == 'alpha':
guess_param = 'a'
else:
guess_param = 'asp'
Guesser.guess(pss, guess_param)
else:
guess_param = None
Guesser.guess(pss, guess_param)
#working on this
def guess(pss, guess_param):
t1 = time.time()
tries = 0
#single passwords
if guess_param != None:
#numeric passwords
if guess_param == 'n':
gd = 0
zeroes = 3
gp = 0
while gp != pss:
gp = ('0' * zeroes) + str(gd)
tries = tries + 1
len1 = len(str(gd))
#adds to the guess digit to cycle through possibilities
gd = gd + 1
len2 = len(str(gd))
if gd > (10 ** len(gp)):
zeroes = zeroes + 1
gd = 0
if len2 > len1:
zeroes = zeroes - 1
#alphanumeric
####PROBLEM AREA###
elif guess_param == 'a':
letters = list(string.digits) + list(string.ascii_letters)
ll = len(letters) - 1
x = -1
gd = [letters[x]]
gdl = 1
xpos = 0
gp = 0
while gp != pss:
xpos = len(gd) - 1
if x + 1 <= 61:
x = x + 1
if xpos == 0:
gd = [x]
else:
gd[xpos] = x
else:
gdl = gdl + 1
x = -1
xpos = len(gd) - 1
gd.insert(0, 0)
for n in range(1, (len(gd) - 1)):
try:
gd[xpos - n] = gd[xpos - n] + 1
break
except gd[xpos - n] + 1 > ll:
gd[xpos - n] = 0
gp = gd
for n in range(0, (len(gp) - 1)):
for item in gp:
gp[n] = letters[item]
gp = ''.join(str(e) for e in gp)
print(gp)
###PROBLEM AREA ABOVE###
#for multiple passwords
else:
#for n in range(0, (len(pass_dict) - 1)):
pass
t2 = time.time()
t = t2 - t1
#simplifying the time it took; not really necessary but whatever
if t/60 > 1:
t = t/60
unit = 'minutes'
elif t/3600 > 1:
t = t/3600
unit = 'hours'
else:
unit = 'seconds'
print('Password guessed! Password was %s\nIt took %s %s and %s tries.' % (gp, t, unit, tries))
答案 0 :(得分：0)
好像你正在尝试创建自己的逻辑来创建alpha数字的所有排列。但是为什么！
Python可以很容易地完成这样的微不足道的事情。
import string
num = string.digits
alphanum = string.digits + string.ascii_lowercase
现在获取3位数字的所有可能数字：
for i in itertools.product(num, num, num):
print("".join(i))
要以通用方式执行此操作，请使用列表乘法运算符。此示例查找包含5位数的所有排列：
for i in itertools.product(*[num] * 5):
print("".join(i))
你可以循环使用它来做1-6个字母的密码。
因此，通过自然延伸，alphanum只是：
for i in itertools.product(*[alphanum] * 5):
print("".join(i))