def函数中的加密代码用python编写

时间:2018-04-26 13:34:39

标签: python validation input

在以下代码中需要一些帮助,因为它进入无限循环并且不验证用户输入:get_offset是函数。刚编辑需要一些帮助,加密部分要在定义的函数中完成

def get_offset(offset):
    while True:
        value = (offset)
        if value < 1:
            print("")
        if value > 94:
            print("")
    return offset

my_details = display_details()

get_choice = get_menu_choice()

print(my_details)
print(get_choice)

count = 10

while count > 0:

    count -=1

    encrypted = ""

    choice = int(input("What would you like to do [1,2,3,4,5]: "))

    while choice <1 and choice > 5:
        print("Invalid choice, please enter either 1, 2, 3, 4 or 5.")

    if choice == 1:
        string_input = input("Please enter string to encrypt: ")
        input_offset = get_offset(int(input("Please enter offset value (1 to 94): ")))

   **for letter in string_input:
        x = ord(letter)
        encrypted += chr(x + input_offset)
        if x < 32:
            x += 94
        if x > 126:
            x -= 94

    print(encrypted)**

1 个答案:

答案 0 :(得分:0)

你必须在内部while循环中分配choice个新值。这样,在反复迭代之前会考虑新的输入。

这应该可以解决问题:

while choice < 1 and choice > 5:
    choice = int(input("Invalid choice, please enter either 1, 2, 3, 4 or 5."))

随机建议:如果您希望数据集严格等于[1,2,3,4,5],则应避免使用choice<1choice>5。这使您的程序容易受到3.5等无效输入的攻击。

相反,你应该这样做:

valid_inputs = [1, 2, 3, 4, 5]

choice = int(input("Pick a number from [1, 2, 3, 4, 5]:"))

while choice not in valid_inputs:
    choice = int(input("Invalid choice, please enter either 1, 2, 3, 4 or 5."))

<rest of stuff here>

我从您的评论中了解到,在get_offset()中,您希望用户输入介于1到94之间的值。如果用户输入的内容超出该范围,您需要一次又一次地提问。下面的代码应该做你想要的:

def get_offset():
   # first, ask for a value.
   offset = int(input("Enter a value between 1 and 94:"))
   while offset < 1 and offset > 94:
      # if the value is invalid, trap user inside this while loop.
      # user will be stuck here (while loop will return true) 
      # until a valid input is received.
      offset = int(input("Invalid input. Please enter a value between 1 and 94:"))
   # if we proceed down here, it means we are over while loop 
   # and it implies that user has given us valid input. return value.
   return offset

其次,在主要的while循环中,您可以像这样调用get_offset()

if choice == 1:
    string_input = input("Please enter string to encrypt: ")
    input_offset = get_offset()