好的，所以我正在写一个python加密代码。代码比它需要的时间长，所以我把它的一部分切换到了函数。一旦我开始这样做，就说它无法找到变量。为了解决这个问题，我将变量设置为全局变量。但是，它仍然无效，并给我错误：
Traceback (most recent call last):
File "C:/Users/Alex/Desktop/Encryptor.py", line 272, in <module>
startup()
File "C:/Users/Alex/Desktop/Encryptor.py", line 257, in startup
encrypt()
File "C:/Users/Alex/Desktop/Encryptor.py", line 44, in encrypt
maincrypt(2)
File "C:/Users/Alex/Desktop/Encryptor.py", line 12, in maincrypt
print newtext
UnboundLocalError: local variable 'newtext' referenced before assignment
这是我的代码：
letters = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
def maincrypt(num):
print otext
print textlist
print newtextlist
print newtext
print times
print i
add = times + num
while add > 25:
add = add - 25
if ord(i) >= 65 and ord(i) <= 90:
newtext += letters[add].uppercase
else:
newtext += letters[add]
def encrypt():
global otext
global textlist
global newtextlist
global newtext
global times
global i
otext = raw_input("Text?")
textlist = []
newtextlist = []
newtext= ""
times = 0
for i in otext:
textlist.append(i)
for i in textlist:
if i == 'a' or i == 'A':
maincrypt(0)
elif i == 'b' or i == 'B':
maincrypt(1)
elif i == 'c' or i == 'C':
maincrypt(2)
elif i == 'd' or i == 'D':
maincrypt(3)
elif i == 'e' or i == 'E':
maincrypt(4)
elif i == 'f' or i == 'F':
maincrypt(5)
elif i == 'g' or i == 'G':
maincrypt(6)
elif i == 'h' or i == 'H':
maincrypt(7)
elif i == 'i' or i == 'I':
maincrypt(8)
elif i == 'j' or i == 'J':
maincrypt(9)
elif i == 'k' or i == 'K':
maincrypt(10)
elif i == 'l' or i == 'L':
maincrypt(11)
elif i == 'm' or i == 'M':
maincrypt(12)
elif i == 'n' or i == 'N':
maincrypt(13)
elif i == 'o' or i == 'O':
maincrypt(14)
elif i == 'p' or i == 'P':
maincrypt(15)
elif i == 'q' or i == 'Q':
maincrypt(16)
elif i == 'r' or i == 'R':
maincrypt(17)
elif i == 's' or i == 'S':
maincrypt(18)
elif i == 't' or i == 'T':
maincrypt(19)
elif i == 'u' or i == 'U':
maincrypt(20)
elif i == 'v' or i == 'V':
maincrypt(21)
elif i == 'w' or i == 'W':
maincrypt(22)
elif i == 'x' or i == 'X':
maincrypt(23)
elif i == 'y' or i == 'Y':
maincrypt(24)
elif i == 'z' or i == 'Z':
maincrypt(25)
else:
newtext += i
times += 1
print newtext
def decrypt():
otext = raw_input("Text?")
textlist = []
newtextlist = []
newtext= ""
times = 0
for i in otext:
textlist.append(i)
for i in textlist:
if i == 'a' or i == 'A':
add = 0 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'b' or i == 'B':
add = 1 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'c' or i == 'C':
add = 2 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'd' or i == 'D':
add = 3 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'e' or i == 'E':
add = 4 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'f' or i == 'F':
add = 5 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'g' or i == 'G':
add = 6 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'h' or i == 'H':
add = 7 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'i' or i == 'I':
add = 8 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'j' or i == 'J':
add = 9 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'k' or i == 'K':
add = 10 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'l' or i == 'L':
add = 11 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'm' or i == 'M':
add = 12 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'n' or i == 'N':
add = 13 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'o' or i == 'O':
add = 14 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'p' or i == 'P':
add = 15 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'q' or i == 'Q':
add = 16 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'r' or i == 'R':
add = 17 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 's' or i == 'S':
add = 18 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 't' or i == 'T':
add = 19 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'u' or i == 'U':
add = 20 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'v' or i == 'V':
add = 21 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'w' or i == 'W':
add = 22 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'x' or i == 'X':
add = 23 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'y' or i == 'Y':
add = 24 - times
while add < 0:
add = add + 25
newtext += letters[add]
elif i == 'z' or i == 'Z':
add = 25 - times
while add < 0:
add = add + 25
newtext += letters[add]
else:
newtext += i
times += 1
print newtext
def startup():
upass = raw_input("Password: ")
if upass == "astrocrypt":
print """Welcome!
1: Encrypt
2: Decrypt
3: Credits
4: Quit"""
try:
choice = input()
except SyntaxError:
print "Oops! That's not a valid number! Try again!"
startup()
except NameError:
print "Oops! That's not a valid number! Try again!"
startup()
if choice == 1:
encrypt()
startup()
elif choice == 2:
decrypt()
startup()
elif choice == 3:
print "Made by Alex Taber"
startup()
elif choice == 4:
pass
else:
print "Unknown number"
startup()
else:
print "Incorrect password!"
startup()
答案 0 :(得分：3)
首先，有非常强大且几乎不可破解的加密算法，其中一些预先包装有python，如hmac
import hmac
def Hash_string(ref):
cypher = hmac.new('you secret word', 'plain-text')
return cypher.hexdigest()
简单易用，更强大。你也可以使用blowfish来获得更高级和更安全的项目，虽然默认情况下没有python，但设置起来很简单。
现在到您的代码，这里是如何在python中使用全局变量
# declare variables as global
global num1
global num2
num2=22 # give initial value to num2
def foo():
# bind variable names to the ones in global scope
global num2
print num2 # output: 22
num2 = 88
print num2 # output: 88
global num1
num1 = num2
def bar():
print num1 # output: 88
print num2 # output 88
foo()
bar()
所以你需要将newtext绑定到声明为gobal的文本
我还建议你重写加密看起来像这样
alpha = ['a'，'b'，'c'，'d'，'e'，'f'，'g'，'h'，'i'，'j'，'k'， 'L'， 'M'， 'N'， '0'， 'p'， 'q'， 'R'， 'S'， 'T'， 'U'， 'v'， 'W'，'× ”， 'Y'， 'Z']
数字= [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21， 22,23,24,25]
def map (ch):
chr = ch.lower()
return numeric[alpha.index(chr)]
这会在数字和字母之间创建一个映射，因此您不需要很长的elif列表
希望这有帮助。
EDIT 您可以使用global关键字访问全局变量，但是如果要修改它们，首先需要绑定到全局变量，然后才能更改它们。
EDIT 更改返回数字[index（chr）]返回数字[alpha.index（chr）] 抱歉这个错误：）
答案 1 :(得分：2)
为了保护程序员免受全局变量的影响，在声明它们与global variablename一起使用之前，不会认为全局变量存在于函数中。在maincrypt中，您不会声明global newtext，也不会通过分配来创建本地newtext。 <=运算符要求左手已存在 - 它尚未存在，因此抛出异常。
顺便说一句，如果你计划在其他任何地方重用这个程序，最好使用类变量而不是全局变量，所以1）如果其他人使用相同的全局变量名称，则一切正常2）你可以有两个实例加密器/解密器代码运行和加密/解密不同的东西，甚至在不同的线程上并行，而不会破坏彼此的全局。
此外，此代码：ord(i.upper()) - ord('A')将为'a'和'A'提供0，为'b'和'B'提供1，为'c'和'C'提供2 ...等等在，所以你不需要如此巨大的if / elif链。 （为什么这有效是因为'A'，'B'，'C'等在ASCII中是连续的，所以'B' - 'A'是1，依此类推）