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
使用python datetime模块为datetime.today()添加6个月的时间增量。
http://docs.python.org/library/datetime.html
你当然要解决Johannes weß提出的问题——你说的6个月是什么意思?
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)
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)
当我需要添加几个月或几年的时间,并且不想导入更多的库时,我就会这样做。
import datetime
__author__ = 'Daniel Margarido'
# Check if the int given year is a leap year
# return true if leap year or false otherwise
def is_leap_year(year):
if (year % 4) == 0:
if (year % 100) == 0:
if (year % 400) == 0:
return True
else:
return False
else:
return True
else:
return False
THIRTY_DAYS_MONTHS = [4, 6, 9, 11]
THIRTYONE_DAYS_MONTHS = [1, 3, 5, 7, 8, 10, 12]
# Inputs -> month, year Booth integers
# Return the number of days of the given month
def get_month_days(month, year):
if month in THIRTY_DAYS_MONTHS: # April, June, September, November
return 30
elif month in THIRTYONE_DAYS_MONTHS: # January, March, May, July, August, October, December
return 31
else: # February
if is_leap_year(year):
return 29
else:
return 28
# Checks the month of the given date
# Selects the number of days it needs to add one month
# return the date with one month added
def add_month(date):
current_month_days = get_month_days(date.month, date.year)
next_month_days = get_month_days(date.month + 1, date.year)
delta = datetime.timedelta(days=current_month_days)
if date.day > next_month_days:
delta = delta - datetime.timedelta(days=(date.day - next_month_days) - 1)
return date + delta
def add_year(date):
if is_leap_year(date.year):
delta = datetime.timedelta(days=366)
else:
delta = datetime.timedelta(days=365)
return date + delta
# Validates if the expected_value is equal to the given value
def test_equal(expected_value, value):
if expected_value == value:
print "Test Passed"
return True
print "Test Failed : " + str(expected_value) + " is not equal to " str(value)
return False
# Test leap year
print "---------- Test leap year ----------"
test_equal(True, is_leap_year(2012))
test_equal(True, is_leap_year(2000))
test_equal(False, is_leap_year(1900))
test_equal(False, is_leap_year(2002))
test_equal(False, is_leap_year(2100))
test_equal(True, is_leap_year(2400))
test_equal(True, is_leap_year(2016))
# Test add month
print "---------- Test add month ----------"
test_equal(datetime.date(2016, 2, 1), add_month(datetime.date(2016, 1, 1)))
test_equal(datetime.date(2016, 6, 16), add_month(datetime.date(2016, 5, 16)))
test_equal(datetime.date(2016, 3, 15), add_month(datetime.date(2016, 2, 15)))
test_equal(datetime.date(2017, 1, 12), add_month(datetime.date(2016, 12, 12)))
test_equal(datetime.date(2016, 3, 1), add_month(datetime.date(2016, 1, 31)))
test_equal(datetime.date(2015, 3, 1), add_month(datetime.date(2015, 1, 31)))
test_equal(datetime.date(2016, 3, 1), add_month(datetime.date(2016, 1, 30)))
test_equal(datetime.date(2016, 4, 30), add_month(datetime.date(2016, 3, 30)))
test_equal(datetime.date(2016, 5, 1), add_month(datetime.date(2016, 3, 31)))
# Test add year
print "---------- Test add year ----------"
test_equal(datetime.date(2016, 2, 2), add_year(datetime.date(2015, 2, 2)))
test_equal(datetime.date(2001, 2, 2), add_year(datetime.date(2000, 2, 2)))
test_equal(datetime.date(2100, 2, 2), add_year(datetime.date(2099, 2, 2)))
test_equal(datetime.date(2101, 2, 2), add_year(datetime.date(2100, 2, 2)))
test_equal(datetime.date(2401, 2, 2), add_year(datetime.date(2400, 2, 2)))
只需创建一个datetime.date()对象,调用add_month(date)来添加一个月,调用add_year(date)来添加一个年。
user417751之前回答的重做。也许不是那么python的方式,但它照顾到不同的月份长度和闰年。在本例中,2012年1月31日+ 1个月= 2012年2月29日。
import datetime
import calendar
def add_mths(d, x):
newday = d.day
newmonth = (((d.month - 1) + x) % 12) + 1
newyear = d.year + (((d.month - 1) + x) // 12)
if newday > calendar.mdays[newmonth]:
newday = calendar.mdays[newmonth]
if newyear % 4 == 0 and newmonth == 2:
newday += 1
return datetime.date(newyear, newmonth, newday)
还有另一个解决方案,希望有人会喜欢:
def add_months(d, months):
return d.replace(year=d.year+months//12).replace(month=(d.month+months)%12)
这个解决方案并不适用于所有情况下的第29、30、31天,所以需要更健壮的解决方案(这已经不是那么好了:)):
def add_months(d, months):
for i in range(4):
day = d.day - i
try:
return d.replace(day=day).replace(year=d.year+int(months)//12).replace(month=(d.month+int(months))%12)
except:
pass
raise Exception("should not happen")