我看过一些答案,但没有一个对我有用,我知道这段代码可能有很多错误,但我想知道的是,如果最后的输入是“是”,如何重新启动代码
Code = input("Do you want to encrypt or decrypt? ")
Code = Code.upper()
Answer = ["ENCRYPT","DECRYPT"]
if Code in Answer:
Plaintext = input("What's your message? ")
Plaintext = Plaintext.upper()
Shift = int(input("What's the shift number? "))
LengthPT = len(Plaintext)
CodeLetter = ""
if Code == ("ENCRYPT"):
for i in range (0,LengthPT):
Pletter = ord(Plaintext[i]) -64
Codeletter = Pletter + Shift
if Codeletter > 26:
Codeletter = Codeletter - 26
Codeascii = chr(Codeletter + 64)
CodeLetter = CodeLetter + Codeascii
elif Code == ("DECRYPT"):
for i in range (0,LengthPT):
Pletter = ord(Plaintext[i]) -64
Codeletter = Pletter - Shift
if Codeletter < 0:
Codeletter = Codeletter + 26
Codeascii = chr(Codeletter + 64)
CodeLetter = CodeLetter + Codeascii
else:
print("Wrong answer.")
if Code == ("ENCRYPT"):
print("Encoded Message =", CodeLetter)
elif Code == ("DECRYPT"):
print("Decoded Message =", CodeLetter)
Answer2 = input("Do you want to restart? (Yes/No): ")
让它成为一个函数。检查答案是否为“是”,然后再次调用函数。递归在Python中是很好的,所以这可以工作。在
有很多方法,但这似乎是最短的路径:
通常,您可能也想
.lower ().strip ()
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