Reputation: 10261
In the code shown below, I need to manipulate the time var in python to display a date/time stamp in python to represent that delay.
For example, when the user enters the delay time in hours, I need to set the jcarddeliver var to update itself with the value of the current date/time + delay.
Also it should update the date var as well. For example, if the date is 24 Feb and time is 15:00 hrs and the delay time is 10 hrs, the jcarddeliver date should change to 25 Feb.
jcarddate = time.strftime("%a %m/%d/%y", time.localtime())
jcardtime = time.strftime("%H:%M:%S", time.localtime())
delay = raw_input("enter the delay: ")
jcarddeliver = ??
I just hope I am making sense.
Upvotes: 1
Views: 4833
Reputation: 11
Try this:
now = time.time()
then = now + 365*24*3600
print time.strftime('%Y-%m-%d', time.localtime(then))
Upvotes: 0
Reputation: 2930
You could try the datetime module, e.g.
import datetime
now = datetime.datetime.now()
delay = float (raw_input ("enter delay (s): "))
dt = datetime.timedelta (seconds=delay)
then = now + dt
print now
print then
Upvotes: 3
Reputation: 59553
The result of time.time()
is a floating point value of the number of seconds since the Epoch. You can add seconds to this value and use time.localtime()
, time.ctime()
and other functions to get the result in various forms:
>>> now = time.time()
>>> time.ctime(now)
'Fri Sep 04 16:19:59 2009' # <-- this is local time
>>> then = now + (10.0 * 60.0 * 60.0) # <-- 10 hours in seconds
>>> time.ctime(then)
'Sat Sep 05 02:19:59 2009'
Upvotes: 1
Reputation: 391818
" i need to set the jcarddeliver var to update itself with the value of the current date/time + delay"
How about reformulating this to
jcarddeliver
should be the current date-time plus the delay.
The "update itself" isn't perfectly sensible.
Try the following:
Try the most obvious way of computing "current date-time plus the delay"
print the result.
Try using localtime()
on this result. What do you get?
Upvotes: 0