def函数中的加密代码以python编写 [英] Encryption code in def function to be written in python

问题描述

在以下代码中需要一些帮助，因为它进入了无限循环并且不验证用户输入:get_offset是函数.刚编辑过的需要在定义的函数中完成加密部分的帮助

need some help in the following code as it goes into infinite loop and does not validate user input: the get_offset is the function. Just edited need some help with the encryption part to be done in a function defined

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)**

推荐答案

您必须在内部while循环内分配choice新值.这样，在一次又一次地迭代之前，将考虑新的输入.

you have to assign choice new value inside inner while loop. this way, new input is considered before iterating again and again.

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."))

一个随机建议:如果要使数据集严格等于[1,2,3,4,5]，则应避免使用choice<1和choice>5.这使您的程序容易受到无效输入(例如3.5)的攻击.

A random advice: If you want to keep dataset strictly equal to [1,2,3,4,5], you should avoid choice<1 and choice>5. This makes your program vulnerable to invalid inputs such as 3.5.

相反，您应该这样做:

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，您希望用户键入一个介于1到94之间的值.如果用户输入的值超出该范围，则您想一次又一次地提问.下面的代码应满足您的要求:

What I understood from your comments is that, in get_offset(), you want user to type a value between 1 and 94. If user enters something out of that range, you want to ask question again and again. The code below should do what you want:

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

第二，在main while循环中，您可以像这样调用get_offset():

Secondly, inside main while loop, you can call get_offset() like this:

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

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