Python - 从日期时间字符串中删除时间
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Python -Remove Time from Datetime String
提问by pythonhunter
I have a date string and want to convert it to the date type:
我有一个日期字符串，想将其转换为日期类型：
I have tried to use datetime.datetime.strptime with the format that I want but it is returning the time with the conversion.
我试图使用 datetime.datetime.strptime 和我想要的格式，但它通过转换返回时间。
when = alldates[int(daypos[0])]
print when, type(when)
then = datetime.datetime.strptime(when, '%Y-%m-%d')
print then, type(then)
This is what the output returns:
这是输出返回的内容：
2013-05-07 <type 'str'>
2013-05-07 00:00:00 <type 'datetime.datetime'>
I need to remove the the time: 00:00:00.
我需要删除时间：00:00:00。
采纳答案by Woody Pride
print then.date()
What you want is a datetime.date object. What you have is a datetime.datetime object. You can either change the object when you print as per above, or do the following when creating the object:
你想要的是一个 datetime.date 对象。你有一个 datetime.datetime 对象。您可以在按上述方式打印时更改对象，也可以在创建对象时执行以下操作：
then = datetime.datetime.strptime(when, '%Y-%m-%d').date()
回答by Ignacio Vazquez-Abrams
>>> print then.date(), type(then.date())
2013-05-07 <type 'datetime.date'>
回答by Dovy
For me, I needed to KEEP a timetime object because I was using UTC and it's a bit of a pain. So, this is what I ended up doing:
对我来说，我需要保留一个 timetime 对象，因为我使用的是 UTC，这有点痛苦。所以，这就是我最终做的：
date = datetime.datetime.utcnow()
start_of_day = date - datetime.timedelta(
hours=date.hour,
minutes=date.minute,
seconds=date.second,
microseconds=date.microsecond
)
end_of_day = start_of_day + datetime.timedelta(
hours=23,
minutes=59,
seconds=59
)
Example output:
示例输出：
>>> date
datetime.datetime(2016, 10, 14, 17, 21, 5, 511600)
>>> start_of_day
datetime.datetime(2016, 10, 14, 0, 0)
>>> end_of_day
datetime.datetime(2016, 10, 14, 23, 59, 59)
回答by nagylzs
To convert a string into a date, the easiest way AFAIK is the dateutil module:
要将字符串转换为日期，AFAIK 最简单的方法是 dateutil 模块：
import dateutil.parser
datetime_object = dateutil.parser.parse("2013-05-07")
It can also handle time zones:
它还可以处理时区：
print(dateutil.parser.parse("2013-05-07"))
>>> datetime.datetime(2013, 5, 7, 1, 12, 12, tzinfo=tzutc())
If you have a datetime object, say:
如果您有一个日期时间对象，请说：
import pytz
import datetime
now = datetime.datetime.now(pytz.UTC)
and you want chop off the time part, then I think it is easier to construct a new object instead of "substracting the time part". It is shorter and more bullet proof:
并且你想砍掉时间部分，那么我认为构造一个新对象比“减去时间部分”更容易。它更短，更防弹：
date_part datetime.datetime(now.year, now.month, now.day, tzinfo=now.tzinfo)
It also keeps the time zone information, it is easier to read and understand than a timedelta substraction, and you also have the option to give a different time zone in the same step (which makes sense, since you will have zero time part anyway).
它还保留时区信息，它比 timedelta 减法更容易阅读和理解，并且您还可以选择在同一步骤中提供不同的时区（这是有道理的，因为无论如何您都会有零时间部分） .
回答by ukrutt
If you need the result to be timezone-aware, you can use the replace()method of datetimeobjects. This preserves timezone, so you can do
如果您需要时区感知结果，您可以使用对象的replace()方法datetime。这会保留时区，所以你可以这样做
>>> from django.utils import timezone
>>> now = timezone.now()
>>> now
datetime.datetime(2018, 8, 30, 14, 15, 43, 726252, tzinfo=<UTC>)
>>> now.replace(hour=0, minute=0, second=0, microsecond=0)
datetime.datetime(2018, 8, 30, 0, 0, tzinfo=<UTC>)
Note that this returns a newdatetime object -- nowremains unchanged.
请注意，这将返回一个新的日期时间对象——now保持不变。
