How do I sort a list chronologically from most recent to last?

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question: ⭐ (sorting)

I am just getting into perl,and have a basic simple question. I have written a script that reads in a directory of logfiles, sorts them in reverse chronological order, and outputs them into an html page.

I cannot figure out how to sort the directory deeper than just the month, which comes first (the files in the dir are all listed by month/day/year). How can I sort this so the most recent days are first?

Originally posted as a Categorized Question.

Comment on How do I sort a list chronologically from most recent to last?

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Re: how do I sort a list chronologically from most recent to last?⭐ by chromatic (Archbishop) on Oct 28, 2000 at 06:19 UTC
This is in perlfaq4, specifically How do I sort an array by (anything)? In this case, you should probably look into making a more advanced data structure with split or using substr in a more complex sort function. Something like this may work: `my @sorted = sort { substr($a, 0, 4) <=> substr($b, 0, 4) } @unsorted;` [download] This code is untested, suitable only as an example of technique, not syntax.	[reply] [d/l]
Re: how do I sort a list chronologically from most recent to last? by ashok (Sexton) on Nov 03, 2000 at 01:30 UTC
You can use shell commands to get the files in chronological order i.e by time stamp and copy those into an array. I assume that only log files are stored in that directory. @files = `ls -lt \| tr -s " " \| cut -f9 -d' '`; #@files finally contains files most recent are first [download]	[reply] [d/l]
Re: how do I sort a list chronologically from most recent to last? by I0 (Priest) on Dec 26, 2000 at 21:40 UTC
`@sorted = map {join'/',(split/-/)[1,2,0]} sort map {join'-',(split'/')[2,0,1]} qw(12/31/2000 12/26/2000 01/01/2001);` [download]	[reply] [d/l]