Python为未指定inpu创建泛型函数

Question

我想能够有这个程序打印出一个答案，说像'对不起，我不认识，如果用户输入1-5以外的任何东西'。你知道吗

import time
rating=input()
if rating == '1':
    time.sleep(.5)
    print('Well, you mustve just had a bad day.')
if rating =='2':
    time.sleep(.5)
    print('Second to last means I wasnt even the best at losing...')
if rating =='3':
    time.sleep(.5)
    print('Atleast thats almost passing.')
if rating =='4':
    time.sleep(.5)
    print('Im offended.')
if rating =='5':
    time.sleep(.5)
    print('Well yeah, obviously.')

Answer 1

利用if、elif和else，如果输入不是1、2、3、4、5，那么代码中的else将打印您想要显示的消息。由于elif，下面的代码也将阻止检查所有其他数字。在代码中，您检查了5个if语句中的所有数字。你知道吗

import time
rating=input()
if rating == '1':
    time.sleep(.5)
    print('Well, you mustve just had a bad day.')
elif rating =='2':
    time.sleep(.5)
    print('Second to last means I wasnt even the best at losing...')
elif rating =='3':
    time.sleep(.5)
    print('Atleast thats almost passing.')
elif rating =='4':
    time.sleep(.5)
    print('Im offended.')
elif rating =='5':
    time.sleep(.5)
    print('Well yeah, obviously.')
else:
    print ('Sorry I dont recognize that input')

再次询问用户，直到输入正确的号码（根据您下面的评论）

while rating:
    if rating == '1':
        time.sleep(.5)
        print('Well, you mustve just had a bad day.')
        break
    elif rating =='2':
        time.sleep(.5)
        print('Second to last means I wasnt even the best at losing...')
        break
    elif rating =='3':
        time.sleep(.5)
        print('Atleast thats almost passing.')
        break
    elif rating =='4':
        time.sleep(.5)
        print('Im offended.')
        break
    elif rating =='5':
        time.sleep(.5)
        print('Well yeah, obviously.')
        break
    else:
        print ('Sorry I dont recognize that input. Enter the rating again.')
        rating=input()
        continue

输出（第二种情况）

6
Sorry I dont recognize that input. Enter the rating again.
6
Sorry I dont recognize that input. Enter the rating again.
5
Well yeah, obviously.