[Answered ]-Python Negative time difference

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1👍

If you want to compute the total hours in a datetime.timedelta object, which is what you get when you compute the difference between to datetime.datetime objects, you can use total_seconds:

td = datetime.datetime.now() - datetime.datetime(2021, 2, 15)

total_hrs = td.total_seconds() // 3600

I don’t get why you can’t do directly self.finish_work - self.start_work ? And also how can that be a negative number? You cannot finish working one day before starting, so I assume, that your code has an issue. Probably in the use of datetime.combine where a job is started at 16:00 on day X and finished at 1:00 on day X + 1. For your code the start and finishing dates will be the same and only the hour will change, giving you the negative difference.

👤dzang

0👍

It turned out to be solved. Added a condition: if self.finish_work < self.start_work: add 24 hours

def hours_work(self):
        if self.finish_work < self.start_work:
            time = datetime.combine(date.today(), self.finish_work) + timedelta(hours=24)
            result = time - datetime.combine(date.today(), self.start_work)
        else:
            result = datetime.combine(date.today(), self.finish_work) - datetime.combine(date.today(), self.start_work)
        hours_work1 = str(result).rsplit(':', 1)[0]
        hours_worked = hours_work1
        return hours_worked
👤lacosst

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