Split Laravel Log Files by Date - bash

Question

I've inherited a Laravel system with a large single log file that is currently around 17GB in size, I'm now rotating future log files monthly, however I need to split the existing log by month.
The date is formatted as yyyy-mm-dd hh:mm:ss ("[2018-06-28 13:32:05]"). Does anybody know how I could perform the split using only bash scripting (e.g. through use of awk, sed etc.).
The input file name is laravel.log. I'd like output files to have format such as laravel-2018-06.log.
Help much appreciated.

Answer 1

Since the information you provide is a bit sparse, I will go with the following assumptions :
each log-entry is a single line
somewhere there is always one string of the form [yyyy-mm-dd hh:mm:ss], if there are more, we take the first.
your log-file is sorted in time.
The regex which matches your date is,
\\[[0-9]{4}(-[0-9]{2}){2} ([0-9]{2}:){2}[0-9]{2}\\]
or a bit less strict
\\[[-:0-9 ]{19}\\]
So we can use this in combination with match(s,ere) to get the desired string :
awk 'BEGIN{ere="\\[[0-9]{4}(-[0-9]{2}){2} ([0-9]{2}:){2}[0-9]{2}\\]"}
{ match($0,ere); fname="laravel-"substr($0,RSTART+1,7)".log" }
(fname != oname) { close(oname); oname=fname }
{ print > oname }' laravel.log
As you say that your file is a bit on the large side, you might want to test this first on a subset which covers a couple of months.
$ head -10000 laravel.log > laravel.head.log
$ awk '{...}' laravel.head.log
$ md5sum laravel.head.log
$ cat laravel.*-*.log | md5sum
If the md5sum is not matching, you might have a problem.