How to list the n'th youngest file (without parsing ls!)

Question

So I know parsing ls is bad!.

Today I found this nugget:

FILENAME=`ls -t $READ_FOLDER | head -1`

So I believe originally this was designed to grab the newest filename, I actually now want to adapt it, to grab maybe the 5th newest file as the process that's checking it is a little slow, and so we can't be sure they have been processed yet.

I'm working in bash, but not opposed to a sh generic solution too.

Answer 1

Perlishly:

#!/usr/bin/env perl

use strict;
use warnings;

my %files_by_mtime;

foreach my $file ( glob ( "./*" ) ) {
   my ( $mtime ) = (stat($file))[9];
   $files_by_mtime{$file} = $mtime; 
}

foreach my $f (  sort { $files_by_mtime{$b} <=> $files_by_mtime{$a} } keys %files_by_mtime ) {
   print "$f $files_by_mtime{$f}\n";
}

Reducing to:

perl -e '%f = map { $_ => (stat)[9] } glob("./*");print "" . ( sort { $f{$b} <=> $f{$a} } keys %f )[4];'

Where the "N"th file is 4, because perl arrays start from zero.

Answer 2

Assuming you have access to GNU tools (you do if you're running Linux), I would use stat instead. For example:

$ ls -l
total 0
-rw-r--r-- 1 terdon terdon 0 Sep 15 16:49 file1
-rw-r--r-- 1 terdon terdon 0 Sep 15 16:39 file2
-rw-r--r-- 1 terdon terdon 0 Sep 15 16:29 file3
-rw-r--r-- 1 terdon terdon 0 Sep 15 16:19 file4
-rw-r--r-- 1 terdon terdon 0 Sep 15 16:09 file5
-rw-r--r-- 1 terdon terdon 0 Sep 15 15:59 file6
-rw-r--r-- 1 terdon terdon 0 Sep 15 15:49 file7

So, the 5th newest file is file5. To print just that, you can do:

$ stat --printf '%Y %n\0' * | sort -zrnk1 | 
    awk -vRS='\0' 'NR==5{sub(/^[^ ]* /,"",$0); print}'

You could then easily make this into a shell function that can take N (5 in your example) as an argument. Just add these lines to your ~/.bashrc or equivalent:

nthfile() {
  stat --printf '%Y %n\0' * | sort -zrnk1 |
     awk -vRS='\0' -vn="$1" 'NR==n{sub(/^[^ ]* /,"",$0); print}'
}

Note that this will also show you directories. If you need it to match hidden files as well run (assuming you're using bash) shopt -s dotglob before the above command.

Explanation

• stat --printf '%Y %n\0' * : for each file or directory in the current folder, print the modification date in seconds since the epoch (%s) and the file name (%n) and end each line with \0 instead of \n. This lets us deal correctly with file names containing newline characters.
• sort -zrnk1 : sort the output in reverse sort order (-r), from the newest to the eldest. The -z tells sort to expect null-terminated input lines. The -n tells it to sort numerically and the -k1 to only consider the first field when sorting.
• awk:
  • -vRS='\0' : set the input record (line) separator to \0;
  • vn=$1 : set the variable n to whatever was given as input to the function;
  • NR==n{} : run this only on line n (5 in the first example);
  • sub(/^[^ ]* /,"",$0); print : substitute all non-space characters from the beginning of the line (^[^ ]*) up to the 1st space and print the result

Answer 3

You can use GNU find to list the files with the modified time expressed as epoch time, then use sort to sort the list, finally head and tail to get the desired numbered file name :

find . -maxdepth 1 -type f -printf '%T@ %p\n' | sort -k1,1nr | head -5 | tail -1

Answer 4

With zsh:

print -r -- *(.Dom[5])

give you the 5th file in current directory sorted by modification time. Change m to a to sort by access time. . and D include only regular and hidden files.

Answer 5

One way to do this (assuming no newlines in your file names) would be

ls -t ... | head -n5 | tail -n1