How to get the most recent file

Question

i am newbie in Python language and i need write a code that list directory that contains files with random names, for example:

JuniperAccessLog-standalone-FCL_VPN-20120319-1110.gz
JuniperAccessLog-standalone-FCL_VPN-20120321-1110.gz

I need get the more recent file

I try this, but without success.

import os
from datetime import datetime 

t = datetime.now()
archive = t.strftime("JuniperAccessLog-standalone-FCL_VPN-%Y%m%d-%H*.gz")
file = os.popen(archive)

Result:

sh: JuniperAccessLog-standalone-FCL_VPN-20120320-10*.gz: command not found

have a possibility the use this logic ?

Answer 1

If you want the most recent file, you could take advantage of the fact that they appear to sort into date time order:

import os

logdir='.' # path to your log directory

logfiles = sorted([ f for f in os.listdir(logdir) if f.startswith('JuniperAccessLog-standalone-FCL_VPN')])

print "Most recent file = %s" % (logfiles[-1],)