4

TASK

I need to write a bash script that would archivd files by month. That is, files with modification dates of 2016-12 would be packed into the archive 2016-12_archive.tar.gz, and files with modification dates of 2017-01 would be packed into 2017-01_archive.tar.gz archive, etc.

Example:

FILE NAME   MODIFICATION DATE
file1.log       2016-12-30       ----> 2016-12_archive.tar.gz
file2.log       2016-12-31       ----> 2016-12_archive.tar.gz
file3.log       2017-01-01       ----> 2017-01_archive.tar.gz
file4.log       2017-01-02       ----> 2017-01_archive.tar.gz
file5.log       2017-01-15       ----> 2017-01_archive.tar.gz

My difficulties

The main problems I have:

  1. How to get the file modification date via bash?

  2. How to process all the files in a given directory (so that they have been archived in an appropriate archive)?

My attempts to solve the problems

  1. I have found two ways to find out the file modification date: date -r $ file +% F and find dir -name filename -printf '% TY-% Tm-% Td \ n'. They both do not work on the computer (OS AIX, I'm not a root). Also I can't get what command 'ls -lc` shows (it doesn't seem to modification date).

  2. I have only one idea: to get all modification file dates in the YYYY-MM format, and then create a list of their unique values. Then for each item in this list find all files with appropriate modification date.

Consolidated attempts

Using istat to get modification date:

$ istat filename
Inode 86741 on device 10/8 File
Protection: rw-r-----
Owner: 6361(user2) Group: 621(norgroup)
Link count: 1 Length 116 bytes

Last updated: 16 февраля 2017 г., 14:25:11 MSK
Last modified: 16 февраля 2017 г., 14:25:11 MSK
Last accessed: 16 февраля 2017 г., 16:08:46 MSK

This is how I can get Last modified value for each file:

for logFile in *.log; do
   mdfDate=$(istat $logFile | grep "Last modified");
   echo $logFile $mdfDate
done

Output:

file1.log Last modified: 30 декабря 2016 г., 14:25:11 MSK
file2.log Last modified: 31 декабря 2016 г., 14:26:11 MSK
file3.log Last modified: 01 января 2017 г., 14:27:11 MSK
file4.log Last modified: 02 января 2017 г., 14:28:11 MSK
file5.log Last modified: 15 января 2017 г., 14:29:11 MSK

So the next step is to extract date in unix format.

For some reason cut doesn't work correct. Awk is too heavy and sophisticated. Maybe sed?

  • @don_crissti AIX has istat which should be a good first step. – terdon Feb 16 '17 at 21:38
  • @terdon, could you please help me, how can I extract modification date from istat output? – Vikora Feb 17 '17 at 9:56
  • I don't have an AIX machine so have no access to istat. It might help if you ran it and added the output you get to your question. – terdon Feb 17 '17 at 10:06
  • @terdon, I added an output of istat. Could you please help me?) – Vikora Feb 20 '17 at 21:41
2

If you had access to GNU date, this would be much easier. As it is, it really would be simpler to just use a more sophisticated language. For example, Perl:

#!/usr/bin/perl -w
use strict;
use POSIX qw(strftime);

my $targetDir = $ARGV[0] || ".";
my %tarFiles;
open(my $input, '-|', "find \"$targetDir\" -type f -name '*.log'");
while (<$input>) {
    # remove trailing newlines
    chomp;
    ## Get the file name
    my $file = $_;
    # Open it as a file handle for stat()
    open(my $fh, '<', "$file") or die;
    # Get the file's stats
    my @stats = stat($fh);
    close($fh);
    # modification time
    my $mtime = $stats[9];
    # Convert to YYYY-MM and build the tar file name
    my $tarfile = strftime "%Y-%m_archive.tar.gz", localtime($mtime);
    # Add to the list of files for this tar file
    push @{$tarFiles{$tarfile}}, qq("$file");
}

for my $tarFile (keys(%tarFiles)) {
    # Build the command that creates the tar file
    my $tarCom = "tar cvzf $tarFile @{$tarFiles{$tarFile}}";
    print "COMMAND: $tarCom\n";

    # Uncomment this line to run the command
    # system("$tarCom")
}                           

Save the script as makeTars.pl (or whatever you like) somewhere in your $PATH, make it executable (chmod +x /path/to/makeTars.pl) and then run like this:

makeTars.pl /path/to/target/dir

For example:

$ ls -l
total 0
-rw-r--r-- 1 terdon terdon 0 Dec 30 00:00  file1.log
-rw-r--r-- 1 terdon terdon 0 Dec 31 00:00  file2.log
-rw-r--r-- 1 terdon terdon 0 Jan  1  2016  file3.log
-rw-r--r-- 1 terdon terdon 0 Jan  2  2016  file4.log
-rw-r--r-- 1 terdon terdon 0 Jan  3  2016  file5.log
-rw-r--r-- 1 terdon terdon 0 Jan  3  2016 'file5 with spaces.log'
$ makeTars.pl .
COMMAND: tar cvzf 2017-02_archive.tar.gz "."
COMMAND: tar cvzf 2016-12_archive.tar.gz "./file2.log" "./file1.log"
COMMAND: tar cvzf 2016-01_archive.tar.gz "./file5 with spaces.log" "./file5.log" "./file4.log" "./file3.log"

Once you're satisfied that it will do what you want, uncomment the last line (system("$tarCom")) to make it actually create the tar files.

NOTE that this will break if your file names contain newlines, but I hope that will not be a problem with log files.

  • 1
    Wow, @terdon, it's awesome, thank you! But I don't know Perl, so I'm starting to learn it right now to understand this code:). – Vikora Feb 21 '17 at 10:14

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