如何增加日期时间?
for i in range(1, 35)
date = datetime.datetime(2003, 8, i)
print(date)
但我需要正确地度过几个月和几年?有什么想法吗?
答案 0 :(得分:196)
date = datetime.datetime(2003,8,1,12,4,5)
for i in range(5):
date += datetime.timedelta(days=1)
print(date)
答案 1 :(得分:46)
使用timedelta对象可以完成递增日期:
import datetime
datetime.datetime.now() + datetime.timedelta(days=1)
在Python文档中查找timedelta对象:http://docs.python.org/library/datetime.html
答案 2 :(得分:7)
这是使用dateutil的relativedelta在日期添加天数的另一种方法。
from datetime import datetime
from dateutil.relativedelta import relativedelta
print 'Today: ',datetime.now().strftime('%d/%m/%Y %H:%M:%S')
date_after_month = datetime.now()+ relativedelta(day=1)
print 'After a Days:', date_after_month.strftime('%d/%m/%Y %H:%M:%S')
<强>输出:强>
今天:25/06/2015 20:41:44
过了一天:01/06/2015 20:41:44
答案 3 :(得分:7)
在某些情况下,所有当前答案都是错误的,因为它们不认为时区相对于UTC会更改其偏移量。因此,在某些情况下,添加24小时不同于添加日历日。
以下解决方案适用于萨摩亚,并保持当地时间恒定。
def add_day(today):
"""
Add a day to the current day.
This takes care of historic offset changes and DST.
Parameters
----------
today : timezone-aware datetime object
Returns
-------
tomorrow : timezone-aware datetime object
"""
today_utc = today.astimezone(datetime.timezone.utc)
tz = today.tzinfo
tomorrow_utc = today_utc + datetime.timedelta(days=1)
tomorrow_utc_tz = tomorrow_utc.astimezone(tz)
tomorrow_utc_tz = tomorrow_utc_tz.replace(hour=today.hour,
minute=today.minute,
second=today.second)
return tomorrow_utc_tz
# core modules
import datetime
# 3rd party modules
import pytz
# add_day methods
def add_day(today):
"""
Add a day to the current day.
This takes care of historic offset changes and DST.
Parameters
----------
today : timezone-aware datetime object
Returns
-------
tomorrow : timezone-aware datetime object
"""
today_utc = today.astimezone(datetime.timezone.utc)
tz = today.tzinfo
tomorrow_utc = today_utc + datetime.timedelta(days=1)
tomorrow_utc_tz = tomorrow_utc.astimezone(tz)
tomorrow_utc_tz = tomorrow_utc_tz.replace(hour=today.hour,
minute=today.minute,
second=today.second)
return tomorrow_utc_tz
def add_day_datetime_timedelta_conversion(today):
# Correct for Samoa, but dst shift
today_utc = today.astimezone(datetime.timezone.utc)
tz = today.tzinfo
tomorrow_utc = today_utc + datetime.timedelta(days=1)
tomorrow_utc_tz = tomorrow_utc.astimezone(tz)
return tomorrow_utc_tz
def add_day_dateutil_relativedelta(today):
# WRONG!
from dateutil.relativedelta import relativedelta
return today + relativedelta(days=1)
def add_day_datetime_timedelta(today):
# WRONG!
return today + datetime.timedelta(days=1)
# Test cases
def test_samoa(add_day):
"""
Test if add_day properly increases the calendar day for Samoa.
Due to economic considerations, Samoa went from 2011-12-30 10:00-11:00
to 2011-12-30 10:00+13:00. Hence the country skipped 2011-12-30 in its
local time.
See https://stackoverflow.com/q/52084423/562769
A common wrong result here is 2011-12-30T23:59:00-10:00. This date never
happened in Samoa.
"""
tz = pytz.timezone('Pacific/Apia')
today_utc = datetime.datetime(2011, 12, 30, 9, 59,
tzinfo=datetime.timezone.utc)
today_tz = today_utc.astimezone(tz) # 2011-12-29T23:59:00-10:00
tomorrow = add_day(today_tz)
return tomorrow.isoformat() == '2011-12-31T23:59:00+14:00'
def test_dst(add_day):
"""Test if add_day properly increases the calendar day if DST happens."""
tz = pytz.timezone('Europe/Berlin')
today_utc = datetime.datetime(2018, 3, 25, 0, 59,
tzinfo=datetime.timezone.utc)
today_tz = today_utc.astimezone(tz) # 2018-03-25T01:59:00+01:00
tomorrow = add_day(today_tz)
return tomorrow.isoformat() == '2018-03-26T01:59:00+02:00'
to_test = [(add_day_dateutil_relativedelta, 'relativedelta'),
(add_day_datetime_timedelta, 'timedelta'),
(add_day_datetime_timedelta_conversion, 'timedelta+conversion'),
(add_day, 'timedelta+conversion+dst')]
print('{:<25}: {:>5} {:>5}'.format('Method', 'Samoa', 'DST'))
for method, name in to_test:
print('{:<25}: {:>5} {:>5}'
.format(name,
test_samoa(method),
test_dst(method)))
Method : Samoa DST
relativedelta : 0 0
timedelta : 0 0
timedelta+conversion : 1 0
timedelta+conversion+dst : 1 1
答案 4 :(得分:5)
这对我来说是一个简单的解决方案:
if DateTime.Now >= January && DateTime.Now < August && DateTime.Now >= 9am && DateTime.Now < 12pm
//do something
答案 5 :(得分:1)
您还可以导入timedelta,以使代码更简洁。
from datetime import datetime, timedelta
date = datetime.now() + timedelta(seconds=[delta_value])
然后将日期转换为字符串
date = date.strftime('%Y-%m-%d %H:%M:%S')
Python一个衬里是
date = (datetime.now() + timedelta(seconds=[delta_value])).strftime('%Y-%m-%d %H:%M:%S')
答案 6 :(得分:1)
最简单的解决方案
from datetime import timedelta, datetime
date = datetime(2003,8,1,12,4,5)
for i in range(5):
date += timedelta(days=1)
print(date)
答案 7 :(得分:0)
一个根本没有库的简短解决方案。 :)
d = "8/16/18"
day_value = d[(d.find('/')+1):d.find('/18')]
tomorrow = f"{d[0:d.find('/')]}/{int(day_value)+1}{d[d.find('/18'):len(d)]}".format()
print(tomorrow)
# 8/17/18
请确保“ 字符串d ”的格式实际上为%m/%d/%Y
,这样您就不会从一个月过渡到下一个月