我得到了一个像这样的 start_date：
from django.utils.timezone import utc
import datetime
start_date = datetime.datetime.utcnow().replace(tzinfo=utc)
end_date = datetime.datetime.utcnow().replace(tzinfo=utc)
duration = end_date - start_date
我得到的输出结果如下：
datetime.timedelta(0, 5, 41038)
我该如何将它转换为正常的时间格式，例如：
10分钟，1小时之类的形式
虽然timedelta对象没有内置的格式化程序，但自己编写格式化程序非常容易：
days, seconds = duration.days, duration.seconds
hours = days * 24 + seconds // 3600
minutes = (seconds % 3600) // 60
seconds = seconds % 60
或者等价地，如果您使用的是 Python 2.7+ 或 3.2+ 版本：
seconds = duration.total_seconds()
hours = seconds // 3600
minutes = (seconds % 3600) // 60
seconds = seconds % 60
'{} minutes, {} hours'.format(minutes, hours)
def convert_timedelta(duration):
days, seconds = duration.days, duration.seconds
hours = days * 24 + seconds // 3600
minutes = (seconds % 3600) // 60
seconds = (seconds % 60)
return hours, minutes, seconds
td = datetime.timedelta(2, 7743, 12345)
hours, minutes, seconds = convert_timedelta(td)
print '{} minutes, {} hours'.format(minutes, hours)
这将打印：
9 minutes, 50 hours
print '{} minute{}, {} hour{}'.format(minutes, 's' if minutes != 1 else '',
hours, 's' if minutes != 1 else '')
或者你可能想编写一个english_plural函数来代替重复自己的's'，而不是手动添加。
从您的评论中可以看出，您实际上希望保持每天分开。这更容易：
def convert_timedelta(duration):
days, seconds = duration.days, duration.seconds
hours = seconds // 3600
minutes = (seconds % 3600) // 60
seconds = (seconds % 60)
return days, hours, minutes, seconds
如果您想将此转换为单个值以存储在数据库中，然后将该单个值转换回格式，请执行以下操作：
def dhms_to_seconds(days, hours, minutes, seconds):
return (((days * 24) + hours) * 60 + minutes) * 60 + seconds
def seconds_to_dhms(seconds):
days = seconds // (3600 * 24)
hours = (seconds // 3600) % 24
minutes = (seconds // 60) % 60
seconds = seconds % 60
return days, hours, minutes, seconds
因此，将其综合起来：
def store_timedelta_in_database(thingy, duration):
seconds = dhms_to_seconds(*convert_timedelta(duration))
db.execute('INSERT INTO foo (thingy, duration) VALUES (?, ?)',
thingy, seconds)
db.commit()
def print_timedelta_from_database(thingy):
cur = db.execute('SELECT duration FROM foo WHERE thingy = ?', thingy)
seconds = int(cur.fetchone()[0])
days, hours, minutes, seconds = seconds_to_dhms(seconds)
print '{} took {} minutes, {} hours, {} days'.format(thingy, minutes, hours, days)
[D day[s], ][H]H:MM:SS[.UUUUUU]的字符串，那么就不需要定制辅助函数。 timedelta对象支持str()操作可以做到这一点。它甚至可以在Python 2.6中运行。>>> from datetime import timedelta
>>> timedelta(seconds=90136)
datetime.timedelta(1, 3736)
>>> str(timedelta(seconds=90136))
'1 day, 1:02:16'
datetime.timedelta 表示两个日期之间的差异，而不是一个日期本身。它仅以天、秒和微秒的形式表示，因为较大的时间单位（如月份和年份）无法清晰地分解（例如30天是1个月还是0.9677个月？）。
如果您想将 timedelta 转换为小时和分钟，则可以使用 total_seconds() 方法获取总秒数，然后进行一些计算：
x = datetime.timedelta(1, 5, 41038) # Interval of 1 day and 5.41038 seconds
secs = x.total_seconds()
hours = int(secs / 3600)
minutes = int(secs / 60) % 60
我认为自己计算不是一个好主意。
如果你只想要一个漂亮的输出，可以使用str()函数将其转换成字符串，或者直接使用print()函数。
如果需要进一步使用小时和分钟，可以使用datetime.strptime()解析为datetime对象（并使用datetime.time()方法提取时间部分），例如：
import datetime
delta = datetime.timedelta(seconds=10000)
time_obj = datetime.datetime.strptime(str(delta),'%H:%M:%S').time()
只需使用strftime :)
类似这样的：
my_date = datetime.datetime(2013, 1, 7, 10, 31, 34, 243366, tzinfo=<UTC>)
print(my_date.strftime("%Y, %d %B"))
def timedelta_tuple(timedelta_object):
return timedelta_object.days, timedelta_object.seconds//3600, (timedelta_object.seconds//60)%60
# Try this code
from datetime import timedelta
class TimeDelta(timedelta):
def __str__(self):
_times = super(TimeDelta, self).__str__().split(':')
if "," in _times[0]:
_hour = int(_times[0].split(',')[-1].strip())
if _hour:
_times[0] += " hours" if _hour > 1 else " hour"
else:
_times[0] = _times[0].split(',')[0]
else:
_hour = int(_times[0].strip())
if _hour:
_times[0] += " hours" if _hour > 1 else " hour"
else:
_times[0] = ""
_min = int(_times[1])
if _min:
_times[1] += " minutes" if _min > 1 else " minute"
else:
_times[1] = ""
_sec = int(_times[2])
if _sec:
_times[2] += " seconds" if _sec > 1 else " second"
else:
_times[2] = ""
return ", ".join([i for i in _times if i]).strip(" ,").title()
# Test
>>> str(TimeDelta(seconds=10))
'10 Seconds'
>>> str(TimeDelta(seconds=60))
'01 Minute'
>>> str(TimeDelta(seconds=90))
'01 Minute, 30 Seconds'
>>> str(TimeDelta(seconds=3000))
'50 Minutes'
>>> str(TimeDelta(seconds=3600))
'1 Hour'
>>> str(TimeDelta(seconds=3690))
'1 Hour, 01 Minute, 30 Seconds'
>>> str(TimeDelta(seconds=3660))
'1 Hour, 01 Minute'
>>> str(TimeDelta(seconds=3630))
'1 Hour, 30 Seconds'
>>> str(TimeDelta(seconds=3600*20))
'20 Hours'
>>> str(TimeDelta(seconds=3600*20 + 3000))
'20 Hours, 50 Minutes'
>>> str(TimeDelta(seconds=3600*20 + 3630))
'21 Hours, 30 Seconds'
>>> str(TimeDelta(seconds=3600*20 + 3660))
'21 Hours, 01 Minute'
>>> str(TimeDelta(seconds=3600*20 + 3690))
'21 Hours, 01 Minute, 30 Seconds'
>>> str(TimeDelta(seconds=3600*24))
'1 Day'
>>> str(TimeDelta(seconds=3600*24 + 10))
'1 Day, 10 Seconds'
>>> str(TimeDelta(seconds=3600*24 + 60))
'1 Day, 01 Minute'
>>> str(TimeDelta(seconds=3600*24 + 90))
'1 Day, 01 Minute, 30 Seconds'
>>> str(TimeDelta(seconds=3600*24 + 3000))
'1 Day, 50 Minutes'
>>> str(TimeDelta(seconds=3600*24 + 3600))
'1 Day, 1 Hour'
>>> str(TimeDelta(seconds=3600*24 + 3630))
'1 Day, 1 Hour, 30 Seconds'
>>> str(TimeDelta(seconds=3600*24 + 3660))
'1 Day, 1 Hour, 01 Minute'
>>> str(TimeDelta(seconds=3600*24 + 3690))
'1 Day, 1 Hour, 01 Minute, 30 Seconds'
>>> str(TimeDelta(seconds=3600*24*2))
'2 Days'
>>> str(TimeDelta(seconds=3600*24*2 + 9999))
'2 Days, 2 Hours, 46 Minutes, 39 Seconds'
def format_timedelta(td):
hours, remainder = divmod(td.total_seconds(), 3600)
minutes, seconds = divmod(remainder, 60)
hours, minutes, seconds = int(hours), int(minutes), int(seconds)
if hours < 10:
hours = '0%s' % int(hours)
if minutes < 10:
minutes = '0%s' % minutes
if seconds < 10:
seconds = '0%s' % seconds
return '%s:%s:%s' % (hours, minutes, seconds)
解决方法之一是从时间差转换为秒开始创建一个相对时间。可以使用time.gmtime(...)方法将秒转换成纪元以来的秒数：
>>> time.strftime("%H:%M:%S",time.gmtime(36901)) # secs = 36901
'10:15:01'
就是这样了！（注：这里有一个链接，可以查看time.strftime()的格式说明符，以便将差异截断为任何单位，根据需要进行调整。...）
值得注意的是，这种技术也是判断当前时区是否处于夏令时的好方法。（它提供了0或1小时的偏移量，基本上可以解释为布尔值。）
import datetime
import pytz
import time
pacific=pytz.timezone('US/Pacific')
now=datetime.datetime.now()
# pacific.dst(now).total_seconds() yields 3600 secs. [aka 1 hour]
time.strftime("%-H", time.gmtime(pacific.dst(now).total_seconds()))
'1'
is_standard_time(...)，其中1表示真，0表示假。您想以那种格式打印日期吗？这是Python文档：http://docs.python.org/2/library/datetime.html#strftime-strptime-behavior
>>> a = datetime.datetime(2013, 1, 7, 10, 31, 34, 243366)
>>> print a.strftime('%Y %d %B, %M:%S%p')
>>> 2013 07 January, 31:34AM
对于时间差：
>>> a = datetime.timedelta(0,5,41038)
>>> print '%s seconds, %s microseconds' % (a.seconds, a.microseconds)
但请注意，您应确保它具有相关的值。对于上述情况，它没有小时和分钟值，您应该从秒钟开始计算。
datetime.timedelta(hours=1, minutes=10)
#python 2.7