Question

I am making a search script for parsing log files in archives and am stuck at one particular thing I didnt find resources anywhere else.

I need to somehow replace the “id1” of this string for “anything”. Simply put I want the script to take anything in that particular place in path. Is this possible only by using logZip as one huge regex?

logZip = ("c:/logs/" + str(id1) + "/" + str(id2) + "/" + str(id1) + "_log_" + str(dateID) + "_" + str(hourID) + "-00_" + str(id2) + ".zip")

As requested, Im adding a sample paths and files I want to define:

c:/logs/aaa1/sss2/aaa1_log_2013-01-14-13-00_sss2.zip
c:/logs/aaa2/sss2/aaa2_log_2013-01-14-13-00_sss2.zip
c:/logs/aaa3/sss2/aaa3_log_2013-01-14-13-00_sss2.zip

Next step will be to make the same thing with sss2 part.

Answer 1

import glob
import os

id1 = '*'
logZip = glob.glob(os.path.join("c:/logs", id1, id2, '{}_log_{}_{}-00_{}.zip'.format(id1, dateID, hourID, id2)))

returns a list of complete paths to files where id1 is replaced by a wildcard *.

os.path.join makes the path construction clearer.

If logZip has to be a single file, you can use glob.iglob, which returns an iterator:

id1 = '*'
mask = os.path.join("c:/logs", id1, id2, '{}_log_{}_{}-00_{}.zip'.format(id1, dateID, hourID, id2))
for logZip in glob.iglob(mask):
    # logZip is a single file