python问题：返回某一天是星期几？

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python问题：返回某一天是星期几？

def date_to_week(year,month,day):
    if year<1990:a=max(1990,year)
    else:a=max(1990,year)+1
    b,count_day=min(1990,year),day
    for j in range(b,a):
        if j % 4 == 0 and j% 100 != 0 or j % 400 == 0:two = 29
        else:two = 28
        list_month=[31,two,31,30,31,30,31,31,30,31,30,31]
        if  year<1990 and j==year:count_day=sum(list_month[:month-1:-1])+list_month[month-1]-count_day
        else:
            if j < a and j != year:mm = 13
            else:mm = month
            count_day += sum(list_month[:mm - 1])
        if j == 1989: count_day = -count_day
    week_day=['日','一','二','三','四','五','六']
    index=count_day%7
    return '星期{}'.format(week_day[index])

以上是编程代码，此处是使用1990.1.1为基准日，基准日可以随便定义，但是一定要知道定义的基准日是星期几。通过判断闰年来确定2月的天数，从而确定该年份是多少天，与基准日求差值，大于基准日选择正值，小于选择负值，再用对应的差值对7求余，剩的余数就是对应周列表的索引。
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                        版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。

原文链接：https://blog.csdn.net/xihuanyigeren2017/article/details/125080051

def date_to_week(year,month,day):
    if year<1990:a=max(1990,year)
    else:a=max(1990,year)+1
    b,count_day=min(1990,year),day
    for j in range(b,a):
        if j % 4 == 0 and j% 100 != 0 or j % 400 == 0:two = 29
        else:two = 28
        list_month=[31,two,31,30,31,30,31,31,30,31,30,31]
        if  year<1990 and j==year:count_day=sum(list_month[:month-1:-1])+list_month[month-1]-count_day
        else:
            if j < a and j != year:mm = 13
            else:mm = month
            count_day += sum(list_month[:mm - 1])
        if j == 1989: count_day = -count_day
    week_day=['日','一','二','三','四','五','六']
    index=count_day%7
    return '星期{}'.format(week_day[index])

                        版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。

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原文链接：https://blog.csdn.net/xihuanyigeren2017/article/details/125080051

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