How to convert current date to epoch timestamp?

Use strptime to parse the time, and call time() on it to get the Unix timestamp.

Assuming you are using a 24 hour time format:

import time;
t = time.mktime(time.strptime("29.08.2011 11:05:02", "%d.%m.%Y %H:%M:%S"));

import time
def expires():
    '''return a UNIX style timestamp representing 5 minutes from now'''
    return int(time.time()+300)

Your code will behave strange if ‘TZ’ is not set properly, e.g. ‘UTC’ or ‘Asia/Kolkata’

So, you need to do below

>>> import time, os
>>> d='2014-12-11 00:00:00'
>>> p='%Y-%m-%d %H:%M:%S'
>>> epoch = int(time.mktime(time.strptime(d,p)))
>>> epoch
1418236200
>>> os.environ['TZ']='UTC'
>>> epoch = int(time.mktime(time.strptime(d,p)))
>>> epoch
1418256000

if you want UTC try some of the gm functions:

import time
import calendar

date_time = '29.08.2011 11:05:02'
pattern = '%d.%m.%Y %H:%M:%S'
utc_epoch = calendar.timegm(time.strptime(date_time, pattern))
print utc_epoch

I think this answer needs an update and the solution would go better this way.

from datetime import datetime

datetime.strptime("29.08.2011 11:05:02", "%d.%m.%Y %H:%M:%S").strftime("%s")

or you may use datetime object and format the time using %s to convert it into epoch time.

from time import time
>>> int(time())
1542449530


>>> time()
1542449527.6991141
>>> int(time())
1542449530
>>> str(time()).replace(".","")
'154244967282'

But Should it not return ?

'15424495276991141'

Short-hand to convert python date/datetime to Epoch (without microseconds)

int(current_date.strftime("%s")) # 2020-01-14  ---> 1578956400
int(current_datetime.strftime("%s")) # 2020-01-14 16:59:30.251176 -----> 1579017570