Question

在下面的代码中，您会看到它要求“移位”值。我的问题是我想将输入限制为1到26。

    For char in sentence:
            if char in validLetters or char in space: #checks for
                newString += char                     #useable characters
        shift = input("Please enter your shift (1 - 26) : ")#choose a shift
        resulta = []
        for ch in newString:
            x = ord(ch)      #determines placement in ASCII code
            x = x+shift      #applies the shift from the Cipher
            resulta.append(chr(x if 97 <= x <= 122 else 96+x%122) if ch != \
            ' ' else ch) # This line finds the character by its ASCII code

我该如何轻松地做到这一点？

Answer 1

使用while循环继续询问他们输入，直到您收到您认为有效的内容：

shift = 0
while 1 > shift or 26 < shift:
    try:
        # Swap raw_input for input in Python 3.x
        shift = int(raw_input("Please enter your shift (1 - 26) : "))
    except ValueError:
        # Remember, print is a function in 3.x
        print "That wasn't an integer :("

如果您获得try-except（例如，如果他们输入int()），您还希望在ValueError来电时设置a屏蔽。< / p>

请注意，如果您使用Python 2.x，则需要使用raw_input()而不是input()。后者将尝试将输入解释为Python代码 - 这可能非常糟糕。

Answer 2

另一种实施方式：

shift = 0
while not int(shift) in range(1,27):
    shift = input("Please enter your shift (1 - 26) : ")#choose a shift

Answer 3

while True:
     result = raw_input("Enter 1-26:")
     if result.isdigit() and 1 <= int(result) <= 26:
         break;
     print "Error Invalid Input"

#result is now between 1 and 26 (inclusive)

Answer 4

尝试这样的事情

acceptable_values = list(range(1, 27))
if shift in acceptable_values:
    #continue with program
else:
    #return error and repeat input

可以放入while循环但是你应该限制用户输入，这样它就不会变成无限

Answer 5

使用if条件：

if 1 <= int(shift) <= 26:
   #code
else:
   #wrong input

或if-condition的while循环：

shift = input("Please enter your shift (1 - 26) : ")
while True:
   if 1 <= int(shift) <= 26:
      #code
      #break or return at the end
   shift = input("Try Again, Please enter your shift (1 - 26) : ")

将用户输入限制为Python中的范围

5 个答案: