Question

Python的新手，我正在编写一个用户输入月份的程序（以数字而非单词 - 例如“3”而不是“March”）和年份（如“2014”） “）。我希望程序显示输入月份和年份的天数。因此，如果用户输入第3个月和2005年，则应显示：

2005年3月有31天。

这是我的代码：

def enteredMonth():
    month = int(input("Enter a month in terms of a number: "))
    return month

def enteredYear():
    year = int(input("Enter a year: "))
    return int(year)

def leapYear(year):
    if year % 4 == 0:
        return True
    else:
        return False

def numberOfDays(month):
    if enteredMonth() == 1:
        month = "January"
        print ("31")
    elif enteredMonth() == 2:
        month = "February"
        print ("28")
    elif enteredMonth() == 2 and leapYear() == true:
        month = "February"
        print ("29")
    elif enteredMonth() == 3:
        month = "March"
        print ("31")
    elif enteredMonth() == 4:
        month = "April"
        print ("30")
    elif enteredMonth() == 5:
        month = "May"
        print ("31")
    elif enteredMonth() == 6:
        month = "June"
        print ("30")
    elif enteredMonth() == 7:
        month = "July"
        print ("31")
    elif enteredMonth() == 8:
        month = "August"
        print ("31")
    elif enteredMonth() == 9:
        month = "September"
        print ("30")
    elif enteredMonth() == 10:
        month = "October"
        print ("31")
    elif enteredMonth() == 11:
        month = "November"
        print ("30")
    elif enteredMonth() == 12:
        month = "December"
        print ("31")
    else:
        print("Please enter a valid month")

def main():
   enteredMonth()
   enteredYear()
   leapYear(year)
   numberOfDays(month)
   print(month, enteredYear(), "has", numberOfDays(month) , "days")

if __name__ == '__main__': 
   main()

问题在于，我没有获得正确的格式，而是获得了类似的内容：

2005年没有天。

请帮忙！我非常感激。

Answer 1

################################################################
# OR Another way to achieve same thing
def numberOfDays(month, year):
    daysInMonths = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
    if month > len(daysInMonths) or year < 0 or month < 0:
        return "Please enter a valid month/year"

    # Here, only divisible by 4 as leap-year but there are many 
    # more conditions for a leap year which you can add in expression
    return daysInMonths[month-1] + int((year % 4) == 0 and month == 2)

def main():
   year = int(input("Enter a year: "))
   month = int(input("Enter a month in terms of a number: "))
   print(month, year, "has", numberOfDays(month, year) , "days")

if __name__ == '__main__': 
   main()

################################################################
# OR using standard library
from calendar import monthrange

def main():
   year = int(input("Enter a year: "))
   month = int(input("Enter a month in terms of a number: "))
   if year < 0 or month < 0:
    print "Pleae enter a valid month/year"
    return
   print(month, year, "has", monthrange(year, month)[1] , "days")

if __name__ == '__main__': 
   main()

Answer 2

在函数numberOfDays()内，您需要返回天数，而不是仅仅打印它。由于它不返回任何内容，因此会在None中打印main()。即您需要在numberOfDays()中包含此行：

return num_days

其中num_days设置为给定月份的天数。事实上，numberOfDays()不应该打印任何东西。只需即可返回所需的天数。

另外，为什么你需要在enteredMonth()内拨打这么多numberOfDays()？您只需检查输入参数month。

在main()内，enteredMonth()和enteredYear()不会设置任何变量。您需要执行month = enteredMonth()。

之类的操作

Answer 3

我不同意YS-L的回答。我正在提交一个以更好的格式显示我的意思。

def enteredMonth():
    month = int(input("Enter a month in terms of a number: "))
    return month

def enteredYear():
    year = int(input("Enter a year: "))
    return int(year)

def leapYear(year):
    if year % 4 == 0:
        return True
    else:
        return False

def numberOfDays(month):
    if month == 1:
        return 31
    elif month == 2:
        return 28
    # etc. Actually, the Pythonic implementation would be to use a dict
    # that maps month numbers to their usual number of days.

def main():
   month = enteredMonth()
   year = enteredYear()
   is_leapyear = leapYear(year)
   number_of_days = numberOfDays(month)
   print(month, year, "has", number_of_days, "days")

if __name__ == '__main__': 
   main()

这有明显的问题 - 例如，闰年调整从未发生过。但它演示了如何从函数中捕获返回值，我认为这是你正在努力的主要观点。

作为一个更高级的主题，以下是我使用datetime内置库执行此操作的方法：

import datetime

def enteredMonth():
    month = int(input("Enter a month in terms of a number: "))
    return month

def enteredYear():
    year = int(input("Enter a year: "))
    return year

def numberOfDays(year, month):
    first_of_entered_month = datetime.date(year, month, 1)
    someday_next_month = first_of_entered_month + datetime.timedelta(days=31)
    first_of_next_month = someday_next_month.replace(day=1)
    last_of_entered_month = first_of_next_month - datetime.timedelta(days=1)
    return last_of_entered_month.day

def main():
    month = enteredMonth()
    year = enteredYear()
    number_of_days = numberOfDays(year, month)
    pretty_month_year = datetime.date(year, month, 1).strftime('%B %Y')
    print(pretty_month_year, "has", number_of_days, "days")

在Python中转换一个月的天数？

3 个答案: