Python 2.7 如何在Python中查找n天前的日期?
各位晚上好 我想写一个脚本,其中我给python一些天(我们称之为d),它给我d天前的日期 我正在努力学习模块datetime:Python 2.7 如何在Python中查找n天前的日期?,python-2.7,date,datetime,timedelta,Python 2.7,Date,Datetime,Timedelta,各位晚上好 我想写一个脚本,其中我给python一些天(我们称之为d),它给我d天前的日期 我正在努力学习模块datetime: import datetime tod = datetime.datetime.now() d = timedelta(days = 50) a = tod - h Type Error : unsupported operand type for - : "datetime.timedelta" and "datetime.datetime" 感谢您的
import datetime
tod = datetime.datetime.now()
d = timedelta(days = 50)
a = tod - h
Type Error : unsupported operand type for - : "datetime.timedelta" and
"datetime.datetime"
感谢您的帮助如果您的变量有问题,可以从
datetime.datetime.now()中减去timedeltad
,没有问题:
import datetime
tod = datetime.datetime.now()
d = datetime.timedelta(days = 50)
a = tod - d
print(a)
2014-12-13 22:45:01.743172
如果你的论点是,昨天,2天前,3个月前,2年前。下面的函数有助于获取参数的确切日期。
首先需要导入以下日期utils
import datetime
from dateutil.relativedelta import relativedelta
然后实现下面的功能
def get_past_date(str_days_ago):
TODAY = datetime.date.today()
splitted = str_days_ago.split()
if len(splitted) == 1 and splitted[0].lower() == 'today':
return str(TODAY.isoformat())
elif len(splitted) == 1 and splitted[0].lower() == 'yesterday':
date = TODAY - relativedelta(days=1)
return str(date.isoformat())
elif splitted[1].lower() in ['hour', 'hours', 'hr', 'hrs', 'h']:
date = datetime.datetime.now() - relativedelta(hours=int(splitted[0]))
return str(date.date().isoformat())
elif splitted[1].lower() in ['day', 'days', 'd']:
date = TODAY - relativedelta(days=int(splitted[0]))
return str(date.isoformat())
elif splitted[1].lower() in ['wk', 'wks', 'week', 'weeks', 'w']:
date = TODAY - relativedelta(weeks=int(splitted[0]))
return str(date.isoformat())
elif splitted[1].lower() in ['mon', 'mons', 'month', 'months', 'm']:
date = TODAY - relativedelta(months=int(splitted[0]))
return str(date.isoformat())
elif splitted[1].lower() in ['yrs', 'yr', 'years', 'year', 'y']:
date = TODAY - relativedelta(years=int(splitted[0]))
return str(date.isoformat())
else:
return "Wrong Argument format"
然后可以像这样调用函数:
print get_past_date('5 hours ago')
print get_past_date('yesterday')
print get_past_date('3 days ago')
print get_past_date('4 months ago')
print get_past_date('2 years ago')
print get_past_date('today')
下面的代码应该可以工作
from datetime import datetime, timedelta
N_DAYS_AGO = 5
today = datetime.now()
n_days_ago = today - timedelta(days=N_DAYS_AGO)
print today, n_days_ago
h
应该是什么?可能重复的