在运行代码时循环不起作用

Question

def chineseZodiac(year):    
    if (year - 2000) % 12 == 0:
       sign = 'Dragon'
    elif (year - 2000) % 12 == 1:
       sign = 'Snake'
    elif (year - 2000) % 12 == 2:
       sign = 'Horse'
    elif (year - 2000) % 12 == 3:
       sign = 'sheep'
    elif (year - 2000) % 12 == 4:
       sign = 'Monkey'
    elif (year - 2000) % 12 == 5:
       sign = 'Rooster'
    elif (year - 2000) % 12 == 6:
       sign = 'Dog'
    elif (year - 2000) % 12 == 7:
       sign = 'Pig'
    elif (year - 2000) % 12 == 8:
       sign = 'Rat'
    elif (year - 2000) % 12 == 9:
       sign = 'Ox'
    elif (year - 2000) % 12 == 10:
       sign = 'Tiger'
    else:
       sign = 'Hare'
    return sign
year = int(input("enter year:"))
while (year <= 1980 and year >= 2014):
    print("your chinese zodiac is ", chineseZodiac(year))

``没有错误，但是在运行代码时不会显示while循环下的打印... 请帮助我，我是菜鸟:-('

Answer 1

以我的评论为基础。问题的直接原因是没有任何年份可以小于 1980 年且大于 2014 年。另外因为您有一个 while 循环，所以我认为您想要不断地请求一年，直到输入有效的年份

def chineseZodiac(year):
        
    signs = {
        0: 'Dragon',
        1: 'Snake',
        2: 'Horse',
        3: 'sheep',
        4: 'Monkey',
        5: 'Rooster',
        6: 'Dog',
        7: 'Pig',
        8: 'Rat',
        9: 'Ox',
        10: 'Tiger',
        11: 'Hare'
    }
    
    return signs[( year - 2000 ) % 12]

year = 0

in_range = lambda year: (1980 <= year <= 2014)
while not in_range(year):
    year = int(input("enter year:"))
    
    if in_range(year):
        print("your chinese zodiac is", chineseZodiac(year))

Answer 2

您输入年份值，然后检查它是否在可接受的时间范围内。好的。但如果您使用 while 循环执行此操作，您将永远一次又一次地重新打印输出。

只需使用一次性 if 语句，如下所示：

if year >= 1980 and year <= 2014:
    print(f"your chinese zodiac is {chineseZodiac(year)}")

另请注意 print 语句中更好的 F-String-Formatting。