计算两个日期的天数差，可以这样做:

import datetime
import math

issuedate = datetime(2019,5,9)   #calculate the issue datetime
current_date = datetime.datetime.now() #calculate the current datetime
diff_date = current_date - issuedate #//calculate the date difference with time also
amount = fine  #you want change

if diff_date.total_seconds() > 0.0:   #its matching your condition
    days = math.ceil(diff_date.total_seconds()/86400)  #calculate days (in 
    one day 86400 seconds)
    deductable_amount = round(amount,2)*days #calclulated fine for all days

因为如果比截止日期多一秒，我们就得收费

其他使用datetime和比较的答案也只适用于时间，没有日期。

例如，要检查现在是早8点还是早8点，我们可以使用:

import datetime

eight_am = datetime.time( 8,0,0 ) # Time, without a date

并随后与:

datetime.datetime.now().time() > eight_am  

返回True

python是最简单的语言，在python中比较日期非常容易，python操作符<，>和==非常适合datetime对象。 每一个在python中都有自己的含义:

<表示日期早于第一个日期 >表示日期延后 ==表示日期与第一天相同 所以，对于你的情况:

import datetime

date = datetime.datetime(2000, 1, 1) # Replace with whatever you want
now = datetime.datetime.now() # You can even find the current date and time using this expression

if date < now:
    print('past')
elif date > now:
    print('future')
else:
    print('present')
# This would print "past"

使用datetime方法和操作符<及其同类。

>>> from datetime import datetime, timedelta
>>> past = datetime.now() - timedelta(days=1)
>>> present = datetime.now()
>>> past < present
True
>>> datetime(3000, 1, 1) < present
False
>>> present - datetime(2000, 4, 4)
datetime.timedelta(4242, 75703, 762105)

计算两个日期的天数差，可以这样做:

import datetime
import math

issuedate = datetime(2019,5,9)   #calculate the issue datetime
current_date = datetime.datetime.now() #calculate the current datetime
diff_date = current_date - issuedate #//calculate the date difference with time also
amount = fine  #you want change

if diff_date.total_seconds() > 0.0:   #its matching your condition
    days = math.ceil(diff_date.total_seconds()/86400)  #calculate days (in 
    one day 86400 seconds)
    deductable_amount = round(amount,2)*days #calclulated fine for all days

因为如果比截止日期多一秒，我们就得收费

datetime。Date (2011, 1,1) < datetime。date(2011, 1,2)返回True。

datetime。Date (2011, 1,1) - datetime。Date(2011, 1,2)将返回datetime.timedelta(-1)。

datetime。日期(2011,1,1)+日期时间。Date(2011, 1,2)将返回datetime.timedelta(1)。

去看医生。