我正在使用datetime Python模块。我希望从当前日期计算6个月的日期。有人能帮我一下吗?
我想从当前日期生成一个6个月后的日期的原因是为了生成一个回顾日期。如果用户在系统中输入数据,系统将有从输入数据之日起6个月的审查日期。
我正在使用datetime Python模块。我希望从当前日期计算6个月的日期。有人能帮我一下吗?
我想从当前日期生成一个6个月后的日期的原因是为了生成一个回顾日期。如果用户在系统中输入数据,系统将有从输入数据之日起6个月的审查日期。
当前回答
PyQt4的QDate类有一个addmonths函数。
>>>from PyQt4.QtCore import QDate
>>>dt = QDate(2009,12,31)
>>>required = dt.addMonths(6)
>>>required
PyQt4.QtCore.QDate(2010, 6, 30)
>>>required.toPyDate()
datetime.date(2010, 6, 30)
其他回答
修改了Johannes Wei在1new_month = 121情况下的答案。这对我来说非常合适。月份可以是正的,也可以是负的。
def addMonth(d,months=1):
year, month, day = d.timetuple()[:3]
new_month = month + months
return datetime.date(year + ((new_month-1) / 12), (new_month-1) % 12 +1, day)
你说6个月是什么意思?
2009-02-13 + 6个月== 2009-08-13?还是2009-02-13 + 6*30天?
import mx.DateTime as dt
#6 Months
dt.now()+dt.RelativeDateTime(months=6)
#result is '2009-08-13 16:28:00.84'
#6*30 days
dt.now()+dt.RelativeDateTime(days=30*6)
#result is '2009-08-12 16:30:03.35'
更多关于mx的信息。DateTime
“Python -dateutil”(外部扩展)是一个很好的解决方案,但你可以使用内置的Python模块(datetime和datetime)来实现它。
我做了一个简短的代码,来解决它(处理年,月和日)
(运行:Python 3.8.2)
from datetime import datetime
from calendar import monthrange
# Time to increase (in months)
inc = 12
# Returns mod of the division for 12 (months)
month = ((datetime.now().month + inc) % 12) or 1
# Increase the division by 12 (months), if necessary (+ 12 months increase)
year = datetime.now().year + int((month + inc) / 12)
# (IF YOU DON'T NEED DAYS,CAN REMOVE THE BELOW CODE)
# Returns the same day in new month, or the maximum day of new month
day = min(datetime.now().day,monthrange(year, month)[1])
print("Year: {}, Month: {}, Day: {}".format(year, month, day))
下面是一个示例,它允许用户决定如何返回一个日期,其中一天大于一个月中的天数。
def add_months(date, months, endOfMonthBehaviour='RoundUp'):
assert endOfMonthBehaviour in ['RoundDown', 'RoundIn', 'RoundOut', 'RoundUp'], \
'Unknown end of month behaviour'
year = date.year + (date.month + months - 1) / 12
month = (date.month + months - 1) % 12 + 1
day = date.day
last = monthrange(year, month)[1]
if day > last:
if endOfMonthBehaviour == 'RoundDown' or \
endOfMonthBehaviour == 'RoundOut' and months < 0 or \
endOfMonthBehaviour == 'RoundIn' and months > 0:
day = last
elif endOfMonthBehaviour == 'RoundUp' or \
endOfMonthBehaviour == 'RoundOut' and months > 0 or \
endOfMonthBehaviour == 'RoundIn' and months < 0:
# we don't need to worry about incrementing the year
# because there will never be a day in December > 31
month += 1
day = 1
return datetime.date(year, month, day)
>>> from calendar import monthrange
>>> import datetime
>>> add_months(datetime.datetime(2016, 1, 31), 1)
datetime.date(2016, 3, 1)
>>> add_months(datetime.datetime(2016, 1, 31), -2)
datetime.date(2015, 12, 1)
>>> add_months(datetime.datetime(2016, 1, 31), -2, 'RoundDown')
datetime.date(2015, 11, 30)
另一种解决方案:计算下个月的天数总和,并将结果添加到当前日期。
import calendar
import datetime
def date_from_now(months):
today = datetime.datetime.today()
month = today.month
year = today.year
sum_days = 0
for i in range(int(months)):
month += 1
if month == 13:
month = 1
year += 1
sum_days += calendar.monthrange(year, month)[1]
return datetime.date.today() + datetime.timedelta(sum_days)
print(date_from_now(12)) # if to day is 2017-01-01, output: 2019-01-01