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I have an application that creates a lot of large log files. I only
want to keep logs for the previous 4 months. I am looking for a way to
list the contents of the log dir and if the create date on the log is
older than 4 months delete it. I have looked at the os.stat() function
to get the created date but I can't figure out how to get the date to
go back to a previos time.

Any help would be appreciated.

import time
#
# 120 days, 24 hours, 60 monutes, 60 seconds
#
fourmonthsago=time.time()-(120*24*60*60)

Compare fourmonthsago value to modified date of files.
Delete those that are less.

Larry Bates


elake wrote:
> I have an application that creates a lot of large log files. I only
> want to keep logs for the previous 4 months. I am looking for a way to
> list the contents of the log dir and if the create date on the log is
> older than 4 months delete it. I have looked at the os.stat() function
> to get the created date but I can't figure out how to get the date to
> go back to a previos time.
>
> Any help would be appreciated.
>

Is there a way to do this for a whole calendar month?

On 25 Oct 2005 10:22:24 -0700, "elake" <> wrote:

>Is there a way to do this for a whole calendar month?
>
Yes, if you can specify exactly what you want to do. E.g.,
today is 2005-10-25. What date would you like as the earliest
to keep for the four months? What if today's date was not
available in the month of four months ago? Do you want to
have cleansed logs always start on the first day of a month?
if so, do you want to go back four prior months, or consider
any current month days as being a month-log even if partial,
and only include 3 prior months? You haven't defined your requirements

if you wanted to go back to the first day of four months back, maybe
you could call that earliest_time and delete all files where after
import os, stat
you remove files where
os.stat(pathtofile)[stat.ST_MTIME] < earliest_time

going back from current time might go something like

>>> import time
>>> y,m = time.localtime()[:2]
>>> y,m
(2005, 10)
>>> y,m = divmod(y*12+m-1-4,12) # -1 for 0..11 months
>>> m+=1 # back to 1..12
>>> y,m
(2005, 6)
>>> time.strptime('%04d-%02d-01'%(y,m), '%Y-%m-%d')
(2005, 6, 1, 0, 0, 0, 2, 152, -1)
>>> earliest_time = time.mktime(time.strptime('%04d-%02d-01'%(y,m), '%Y-%m-%d'))
>>> earliest_time
1117609200.0
>>> time.ctime(earliest_time)
'Wed Jun 01 00:00:00 2005'

Hm ...
Jun, Jul, Aug, Sep, Oct
-4 -3 -2 -1 -0
I guess that guarantees 4 full prior months.

I used strptime to build a complete 9-tuple rather than doing it by hand, which
I'm not sure off hand how to do

There's another module that does date interval addition/subtraction, but it didn't
come with the batteries in my version.

BTW, make sure all your date stuff is using the same epoch base date, in case you
have some odd variant source of numerically encoded dates, e.g., look at

>>> import time
>>> time.localtime(0)
(1969, 12, 31, 16, 0, 0, 2, 365, 0)
>>> time.ctime(0)
'Wed Dec 31 16:00:00 1969'
>>> time.gmtime(0)
(1970, 1, 1, 0, 0, 0, 3, 1, 0)

Regards,
Bengt Richter

If you're on Python 2.4, then consider whether or not you can use a
TimedRotatingLogFileHandler from the logging module to handle this for
you:
http://www.python.org/doc/2.4.1/lib/node344.html

Of course, that only works if defining a "month" as 30 days is
acceptable. If you genuinely need calendar months, then you still need
to do it manually.