difference between two times in seconds

[23rnj3mn32sdjqw]

Hi,

It should be simple - but I am trying to find the difference between two datetimes as an integer in seconds. Using pywin32 build 212, python 2.5, Windows XP Pro, MySQL 5.0.

import time
import datetime
 
def date4string(string):
    return datetime.date(*time.strptime(string, "%Y-%m-%d %H:%M:%S")[0:3])
 
date1="2009-01-01 00:00:00"
date2="2009-01-02 00:00:00"
d1=date4string(date1)
d2=date4string(date2)
diff=d2-d1
print diff

gives >>> 1 day, 0:00:00

what I want is the answer 86400

Sorry for being an idiot. I have looked at time.mktime;

diff1=time.mktime(diff)
print diff1

but that gives

TypeError: argument must be 9-item sequence, not datetime.timedelta

[bvdet]

I suggest using time.mktime().

def comp_dates(d1, d2):
    # Date format: %Y-%m-%d %H:%M:%S
    return time.mktime(time.strptime(d2,"%Y-%m-%d %H:%M:%S"))-\
           time.mktime(time.strptime(d1, "%Y-%m-%d %H:%M:%S"))

>>> date1="2009-01-01 00:00:00"
>>> date2="2009-01-02 00:00:00"
>>> comp_dates(date1, date2)
86400.0
>>> 

[23rnj3mn32sdjqw]

Hi,

Thanks for the response. It doesn't help though becuase I have failed to describe the problem properly.

I a writing a script to create a MySQL table using data from an existing set of large tables.

I am pulling two dates (its slightly more complex than that) from the existing tables, and I want to put the difference in seconds back in the resulting table.

I can handle all the MyQSL stuff - I am just miserably stuck on how to get the difference between the two dates - in seconds as an integer.

Regards,

Ben

[bvdet]

Why did my post not help? You wanted an answer of 86400 and I showed you how to get an answer of 86400.0. Are you not aware that objects can be type cast in Python?

>>> int(comp_dates(date1, date2))
86400
>>> 

[23rnj3mn32sdjqw]

I am not making myself very clear.

I don't have strings at all. I have two dates being returned as dates from a MySQL query. They are in date format. I need to be able to subtract one date from another and get a difference in seconds.

[23rnj3mn32sdjqw]

I have solved the problem by changing the initial query to return unix_timestamp(date) from the database for each date in question rather than the date. Then subtracting one seconds from epoch from the other seconds from epoch.

I can't belive its not possible to just subtract one date from another and return the number of seconds difference though.

[bvdet]

OK, maybe it's me that's a bit slow. You have two datetime.date objects, d1 and d2, right?

def comp_dates2(d1, d2):
    # Date format: %Y-%m-%d %H:%M:%S
    return time.mktime(time.strptime(d2,"%Y-%m-%d"))-\
           time.mktime(time.strptime(d1, "%Y-%m-%d"))
 
print repr(d1)
print repr(d2)
print int(comp_dates2(str(d1), str(d2)))

Output:

>>> datetime.date(2009, 1, 1)
datetime.date(2009, 1, 2)
86400
>>>

Another way:

>>> def comp_dates3(d1, d2):
...     delta = d2-d1
...     return delta.days*86400+delta.seconds
... 
>>> comp_dates3(d1, d2)
86400
>>> 

[IceMan85]

I am not sure that I understood, but a solution can be :
Converting to epoch time, doing all the stuff you need and than reconverting it back ?

I hope it helped, regards.