在Python中添加5天的日期。

I have a date "10/10/11(m-d-y)" and I want to add 5 days to it using a Python script. Please consider a general solution that works on the month ends also.

我有一个日期“10/10/11(m-d-y)”，我想用Python脚本添加5天的时间。请考虑一个通用的解决方案，在月底也可以。

I am using following code:

我使用以下代码:

import re
from datetime import datetime

StartDate = "10/10/11"

Date = datetime.strptime(StartDate, "%m/%d/%y")

print Date -> is printing '2011-10-10 00:00:00'

打印日期->正在打印“2011-10-10 00:00”

Now I want to add 5 days to this date. I used the following code:

现在我想在这个日期加5天。我使用了以下代码:

EndDate = Date.today()+timedelta(days=10)

Which returned this error:

返回这个错误:

name 'timedelta' is not defined

9 个解决方案

#1

The previous answers are correct but it's generally a better practice to do:

前面的答案是正确的，但通常是更好的做法:

import datetime

Then you'll have, using datetime.timedelta:

然后使用datetime。timedelta:

date_1 = datetime.datetime.strptime(start_date, "%m/%d/%y")

end_date = date_1 + datetime.timedelta(days=10)

#2

Import timedelta first.

进口timedelta第一。

from datetime import timedelta

And Date.today() will return today's datetime, may be you want

today .today()将返回今天的datetime，您可能需要它

EndDate = Date + timedelta(days=10)

#3

I guess you are missing something like that:

我猜你漏掉了这样的东西:

from datetime import timedelta

#4

If you happen to already be using pandas, you can save a little space by not specifying the format:

如果您碰巧已经在使用熊猫，您可以通过不指定格式来节省一点空间:

import pandas as pd
startdate = "10/10/2011"
enddate = pd.to_datetime(startdate) + pd.DateOffset(days=5)

#5

Here is another method to add days on date using dateutil's relativedelta.

下面是另一种使用dateutil的相对论delta添加日期天数的方法。

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(days=5)
print 'After 5 Days:', date_after_month.strftime('%d/%m/%Y %H:%M:%S')

Output:

输出:

Today: 25/06/2015 15:56:09

今天:25/06/2015 15:56:09

After 5 Days: 30/06/2015 15:56:09

5天后:30/06/2015 15:56:09。

#6

Here is a function of getting from now + specified days

这是一个从现在开始+指定天数的函数

import datetime

def get_date(dateFormat="%d-%m-%Y", addDays=0):

    timeNow = datetime.datetime.now()
    if (addDays!=0):
        anotherTime = timeNow + datetime.timedelta(days=addDays)
    else:
        anotherTime = timeNow

    return anotherTime.strftime(dateFormat)

Usage:

用法:

addDays = 3 #days
output_format = '%d-%m-%Y'
output = get_date(output_format, addDays)
print output

#7

In order to have have a less verbose code, and avoid name conflicts between datetime and datetime.datetime, you should rename the classes with CamelCase names.

为了具有更少的详细代码，并避免datetime和datetime之间的名称冲突。datetime，您应该用CamelCase名称来重命名类。

from datetime import datetime as DateTime, timedelta as TimeDelta

So you can do the following, which I think it is clear.

你可以做下面的，我觉得很清楚。

date_1 = DateTime.today() 
end_date = date_1 + TimeDelta(days=10)

Also, there would be no name conflict if you want to import datetime later on.

此外，如果您希望稍后导入datetime，则不会出现名称冲突。

#8

If you want add days to date now, you can use this code

如果您想现在添加日期，您可以使用此代码

from datetime import datetime
from datetime import timedelta


date_now_more_5_days = (datetime.now() + timedelta(days=5) ).strftime('%Y-%m-%d')

#9

This is my initial but more complete development for a generic dateadd function in python

这是我对python中的通用dateadd函数的初步但更完整的开发

import datetime
from dateutil.relativedelta import relativedelta

def dateadd(date, part ,value):

    if part=='year':
        result = date + (value * relativedelta(years = 1))
    elif part == 'month':
        result = date + (value * relativedelta(months = 1))
    elif part == 'day':
        result = date + (value * relativedelta(days = 1))
    elif part == 'hour':
        result = date + (value * relativedelta(hours = 1))
    elif part == 'minute':
        result = date + (value * relativedelta(minutes = 1))
    elif part == 'second':
        result = date + (value * relativedelta(seconds = 1))
    return result


date = datetime.datetime.now()
print(date, "now")
print(dateadd(date,'year',2)   ,"year',2", )
print(dateadd(date,'year',-2)  ,"year',-2", )
print(dateadd(date,'month',2)  ,"month',2", )
print(dateadd(date,'month',-2) ,"month',-2", )
print(dateadd(date,'day',2)    ,"day',2", )
print(dateadd(date,'day',-2)   ,"day',-2", )
print(dateadd(date,'hour',2)   ,"hour',2", )
print(dateadd(date,'hour',-2)  ,"hour',-2", )
print(dateadd(date,'minute',2) ,"minute',2", )
print(dateadd(date,'minute',-2),"minute',-2", )
print(dateadd(date,'second',2) ,"second',2", )
print(dateadd(date,'second',-2),"second',-2", )

Result:
2018-03-08 09:04:12.619699 now
2020-03-08 09:04:12.619699 year',2
2016-03-08 09:04:12.619699 year',-2
2018-05-08 09:04:12.619699 month',2
2018-01-08 09:04:12.619699 month',-2
2018-03-10 09:04:12.619699 day',2
2018-03-06 09:04:12.619699 day',-2
2018-03-08 11:04:12.619699 hour',2
2018-03-08 07:04:12.619699 hour',-2
2018-03-08 09:06:12.619699 minute',2
2018-03-08 09:02:12.619699 minute',-2
2018-03-08 09:04:14.619699 second',2
2018-03-08 09:04:10.619699 second',-2

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