python中ascii加密程序出错

Question

我有几个星期的编码，有一个任务要做，需要加密输入到ascii，或解密消息。我不能使用两种代码来加密和解密，这可以通过输入密码的否定版本(因此函数)来实现--我自己获得了加密，但很难将它作为一个包工作。程序需要从用户那里获得一个选项(加密或解密)，然后使用从文本和密钥返回的值来转换主函数中的文本。我已经做了10个小时的研究和改变了很多元素，但似乎不断下降。任何建议都会很棒。运行时出现的错误，我将其放在代码下面。

def main(function, message, passkey):
    #takes value of mode and applies text. then uses key to encrypt or decrypt
    encrypt = ""
    for x in message:
    if x == " ":
        encrypt += " "
    else:
        encrypt += chr((((ord(x) - 65) + passkey % 26) + 65))


def mode():
    # determines either encryption or decryption. 
    func = input("Which mode would you like? E for encryption or D for decryption.\n").upper()
    if func[0] =='E': return 'E'
    elif func[0] == 'D': return 'D'
    else: print("Not a valid option. Please try again")

def text():
    #depending on value of mode, either input a sentance to encrypt or an ecrypted message to decode.
    if function == 'E':
        sentance = input("Please enter a sentance to encrypt.\n").upper()
        if all(x.isalpha or x.isspace() for x in sentance):
            return text
        else: sentance = input("Only uppercase alpha characters and spaces allowed. Try again.\n")
    else:
        return input("Enter coded message for decrypting:\n")


def key():
    #depending on value of mode, enter positve key to encrypt or same key in negative form to decrypt.
    if function == 'E':
        return abs(int(input("Enter passkey: ")))
    elif function == 'D':
        return -abs(int(input("Enter passkey: ")))

function = mode()
message = text()
passkey = key()

    print (message)
    print(main(function, message, passkey))

运行时:Python3.6.2 (v3.6.2:5fd33b5，7月8日，04:57:36) MSC v.1900 64位(AMD64)关于win32类型的“版权”、“信用”或"license()“以获得更多信息。

重新启动:EX6 test.py =您想要哪种模式？E用于加密，D用于解密。请输入一个句子来加密。hello输入passkey: 4 Traceback (最近一次调用)："C:\Users\ninja\AppData\Local\Programs\Python\Python36\ex6 test.py"，第43行，打印( main (函数，消息，密码键)文件"C:\Users\ninja\AppData\Local\Programs\Python\Python36\ex6 test.py"，第6行，main for x in message: TypeError：'function‘object不可迭代 ""“

Answer 1

在您的错误消息中，它表示一个'function' object is not iterable。这是因为您正在返回text，这是一个函数。

要解决这个问题，只需将第28行中的return语句改为返回sentance。您还需要在main中返回加密消息。

def main(function, message, passkey):
    # takes value of mode and applies text. then uses key to encrypt or decrypt
    encrypt = ""
    for x in message:
        if x == " ":
            encrypt += " "
        else:
            encrypt += chr((((ord(x) - 65) + passkey % 26) + 65))
    return encrypt


def mode():
    # determines either encryption or decryption.
    func = input("Which mode would you like? E for "
                 "encryption or D for decryption.\n").upper()
    if func[0] == 'E':
        return 'E'
    elif func[0] == 'D':
        return 'D'
    else:
        print("Not a valid option. Please try again")


def text():
    # depending on value of mode, either input a sentance
    # to encrypt or an ecrypted message to decode.
    if function == 'E':
        sentance = input("Please enter a sentance to encrypt.\n").upper()
        if all(x.isalpha or x.isspace() for x in sentance):
            return sentance
        else:
            sentance = input("Only uppercase alpha characters "
                             "and spaces allowed. Try again.\n")
    else:
        return input("Enter coded message for decrypting:\n")


def key():
    # depending on value of mode, enter positve key to encrypt
    # or same key in negative form to decrypt.
    if function == 'E':
        return abs(int(input("Enter passkey: ")))
    elif function == 'D':
        return -abs(int(input("Enter passkey: ")))


function = mode()
message = text()
passkey = key()

print(message)
print(main(function, message, passkey))