Remove files older than 5 days in UNIX (date in file name, not timestamp)

Question

I want to delete log files which are older than 5 days from a directory. But deletion should not be based on the timestamp of file. It should be based on the name of file. For Example todays date is 07/05/2012 and the directory contains 10 files of names like ABC_20120430.log, ABC_20120429.log, ABC_20120502.log, ABC_20120320.log etc. I want to be able to remove the files by extracting the date from the name of the file.

Answer 1

I think @oHessling almost has it: Don't parse ls, and you can do more in bash:

four_days=$(date -d "4 days ago" +%Y%m%d)
for f in ABC_[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9].log; do
  date=${f#ABC_}
  date=${date%.log}
  (( $date < $four_days )) && rm "$f"
done

Answer 2

Based on date from filename:

THRESHOLD=$(date -d "5 days ago" +%Y%m%d)
ls -1 ABC_????????.log | 
  sed 'h;s/[_.]/ /g;G;s/\n/ /' | 
  while read A DATE B FILE
  do 
     [[ $DATE -le $THRESHOLD ]] && rm -v $FILE
  done

Answer 3

One way using perl:

Content of script.pl:

use warnings;
use strict;
use Time::Local qw/timelocal/;
use File::Spec;

## Process all input files.
while ( my $file = shift @ARGV ) { 

    ## Remove last '\n'.
    chomp $file;

    ## Extract date from file name.
    my ($date) = $file =~ m/.*_([^.]+)/ or next;

    ## Extract year, month and day from date.
    my ($y,$m,$d) = $date =~ m/(\d{4})(\d{2})(\d{2})/ or next;

    ## Get date in seconds.
    my $time = timelocal 0, 0, 0, $d, $m - 1, $y - 1900 or next;

    ## Get date in seconds five days ago.
    my $time_5_days_ago = time - 5 * 24 * 3600;

    ## Substract them, and if it is older delete it and print the
    ## event.
    if ( $time - $time_5_days_ago < 0 ) { 
        unlink File::Spec->rel2abs( $file ) and printf qq[%s\n], qq[File $file deleted];
    }   
}

To test it I create some files:

touch ABC_20120430.log ABC_20120502.log ABC_20120320.log ABC_20120508.log ABC_20120509.log

Check them with ls -1:

ABC_20120320.log                                                                                                                                                                                                                             
ABC_20120430.log                                                                                                                                                                                                                             
ABC_20120502.log                                                                                                                                                                                                                             
ABC_20120508.log                                                                                                                                                                                                                             
ABC_20120509.log                                                                                                                                                                                                                             
script.pl

Run the script like:

perl script.pl *.log

With following output:

File ABC_20120320.log deleted
File ABC_20120430.log deleted
File ABC_20120502.log deleted

Answer 4

What you could do instead is use the fact that your filenames will sort in chronological order. For instance, to keep the last 5 files:

ls ABC_????????.log | head -n-5 | xargs rm