Remove Timestamp From Date String In Python

I have situation where I should ignore the timestamp in a date string. I have tried the below command but with no luck. 'start' variable used below is in AbsTime (Ex: 01MAY2017 11:

Solution 1:

Your directives are slightly off, and you need to capture all the contents (irrespective of what you want to retain).

start_date = datetime.datetime.strptime(start, '%d%b%Y %H:%M').date()
print start_date.strftime('%d%b%Y')
# '01May2017'

Update - Adding complete code below:

import datetime
start = '01MAY2017 11:45'
start_date = datetime.datetime.strptime(start, '%d%b%Y %H:%M').date()
print start_date.strftime('%d%b%Y')
# 01May2017

Solution 2:

https://docs.python.org/2/library/datetime.html

To convert the string back to a datetime we can use strptime()

Example using strptime()

datetime.datetime.strptime('10Apr2017 00:00', '%d%b%Y %H:%M')

In [17]: datetime.datetime.strptime('10Apr2017 00:00', '%d%b%Y %H:%M')
Out[17]: datetime.datetime(2017, 4, 10, 0, 0)

Use strftime to form your datetime object

Example using now() returns an datetime object that we form to a string

datetime.datetime.now().strftime('%d%b%Y')

In [14]: datetime.datetime.now().strftime('%d%b%Y')
Out[14]: '10Apr2017'

Solution 3:

try this

  import datetime 
  start = '01MAY2017 11:45' 
  start_date = datetime.datetime.strptime(start, '%d%b%Y %H:%M')
  print start_date.strftime('%Y-%m-%d')

Solution 4:

Actually My Best Option to use the very same datetime object to subtract hours , if you do not wish to use strings and use the datetime Object itself.

Here's an Example:

In [1]: date.strftime('%c')
Out[2]: 'Wed Jul 10 00:00:00 2019'

In [3]: date = datetime.datetime.utcnow()

In [4]: date.strftime('%c')
Out[5]: 'Wed Jul 10 19:06:22 2019'

In [6]: date = date - datetime.timedelta(hours = date.hour, minutes = date.minute, seconds = date.second) #Removing hours, mins,secs

In [7]: date.strftime('%c') #use 
Out[8]: 'Wed Jul 10 00:00:00 2019'