TypeError：无法比较原始偏移量和偏移量感知时间

Question

我有这个功能来检查一周中的特定时间和日期。它应该从周日晚上 7:00 到周五晚上 8:00 打印 1 并从周五晚上 8:00 到周日晚上 7:00 打印 0我在周五上午 11:00 和下午 2:00 左右检查了函数，我在标题中收到了错误消息。有人可以准确解释错误的含义吗？我怎么可能解决它？它应该总是检查东部标准时间的时间import pytzfrom datetime import datetime, time, dateest = pytz.timezone('EST')Mon = 0Tue = 1Wed = 2Thur = 3Fri = 4 Sat = 5 Sun = 6weekdays = [Mon, Tue, Wed, Thur]edgecases = [Sun, Fri]weekend = [Sat]curr_day = datetime.now(tz=est).date().weekday()curr_time = datetime.now(tz=est).time()def checktime(curr_day, curr_time):    if curr_day in weekdays or (curr_day == Sun and curr_time > time(19,00,tzinfo=est)) or (curr_day == Fri and curr_time < time(20,00,tzinfo=est)):        print(1)    elif curr_day in weekend or (curr_day == Fri and curr_time >= time(20,00,tzinfo=est)) or (curr_day == Sun and curr_time <= time(19,00,tzinfo=est)):        print(0)回溯错误：Traceback (most recent call last):  File ".\testingtime.py", line 73, in <module>    checktime(curr_day, curr_time)  File ".\testingtime.py", line 67, in checktime    if curr_day in weekdays or (curr_day == Sun and curr_time > time(19,00,tzinfo=est)) or (curr_day == Fri and curr_time < time(20,00,tzinfo=est)):TypeError: can't compare offset-naive and offset-aware times

Answer 1

其一，如果您time()在可识别 tz 的日期时间对象上调用该方法，则生成的time对象将不再携带时区信息 - 因为假设没有日期就没有意义。其次，由于您与静态时间相比，您在那里不需要时区。因此，您可以将您的功能简化为

def checktime(curr_day, curr_time):

    if curr_day in weekdays or (curr_day == Sun and curr_time > time(19,00)) or (curr_day == Fri and curr_time < time(20,00)):

        print(1)

    elif curr_day in weekend or (curr_day == Fri and curr_time >= time(20,00)) or (curr_day == Sun and curr_time <= time(19,00)):

        print(0)

Answer 2

这为我排序了：

dt_obj = dt_obj.replace(tzinfo=None)

请参见python – 无法减去原始偏移和偏移感知日期时间 ... 答案 1