PYTHON > les dates

from datetime import datetime
from datetime import timedelta

AFFICHER LA DATE ET L’HEURE

import datetime
date = datetime.datetime.now()
print(str(date))
'2016-11-19 09:38:51.754029'

date.year // année
2016

date.month // mois
11

date.day // jour
19

date.hour // heure
9

date.minute // minutes
38

date.second // secondes
51

—

Deux caractères :

dump it back to string using .strftime():

import datetime
dt = datetime.datetime.strptime("08/Jan/2012:08:00:00", "%d/%b/%Y:%H:%M:%S")
dt.hour
8
dt.strftime("%H")
'08'

You cannot have leading zero for int:

type(time1.hour)
<class 'int'>

That’s why you have no choice but convert time1.hour to str first and then manipulate the format.

@alecxe solution (which seems elegant and right) or simply use something like that:

"{0:0>2}".format(time1.hour)

Ajouter 1 jour :

print datetime.now() + timedelta(days=1)

Soustraire 60 secondes :

print datetime.now() - timedelta(seconds=60)

Ajouter 2 ans :

print datetime.now() + timedelta(days=730)

Autres parameters pour ~~timedelta~~ : days, seconds, microseconds, milliseconds, minutes, hours, weeks

Ajouter 1 jour ET 5 minutes :

print datetime.now() + timedelta(days=1,minutes=5)