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Year of file creation

 
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# 1  
Old 11-09-2004
Year of file creation
How do I know the year of creation of a file modified less than 6 months from now. Since I "know" that the file has been created
in 2004 since ls -l does not show the year of its creation but can I get the same information through a command etc.
From the ls man page:
Quote:
-l Lists in long format, giving mode, ACL indication,
number of links, owner, size in bytes, and time of
last modification for each file. If the file is a spe-
cial file the size field will instead contain the
major and minor device numbers. If the time of last
modification is greater than six months ago, it is
shown in the format `month date year'; files modified
within six months show `month date time'. If the file
is a symbolic link, the pathname of the linked-to file
is printed preceded by `->'.

Am I clear enough??

Chill
enc.
# 2  
Old 11-09-2004
If the file is created in the current year ( i.e 2004 ) it is not showing the year by ' ls -l ' listing

But if it passes 2004 year ... it will show the year of creation or modification time ;

See the following o/p for one of my directory :

drwxr-xr-x 2 bhargav staff 512 Dec 27 2002 CORBA
-rwxr-xr-x 1 bhargav staff 3353 Jan 15 2003 set_env
-rwxrwxrwx 1 bhargav staff 7222 Jan 21 2003 variables.mk
-rw-r--r-- 1 bhargav staff 740 Nov 09 10:41 x



If it is not showing the year in 'ls -l ' listing ... then it is abvious that the file is modified last time in the same year ;
# 3  
Old 11-09-2004
Code:
filetime.pl t.lis | awk 'BEGIN{FS=":"} {print $1}'

Gives you the year of the last time a file was modified
Here is filetime.pl:
Code:
#!/usr/contrib/bin/perl
#^###################################################################
#^ $Header: filetime.pl,v 1.1 97/01/20 10:17:00 http srvr? $
#^###################################################################
#^ PROGRAM DESCRIPTION
#^ -------------------
#^ This program prints the modification dates of files
#^ It uses the following format: filetime.pl filename
#^ It will accept:  filetime.pl filename1 filename2 filename3
#^                  filetime.pl /tmp/file*
#^ The format of the output is: YYYY:MM:DD:hh:mm:ss filename
#^ example:
#^          
#^           1996:11:15:10:54:25 filename
#^           
############################################
# loop
############################################
while ($curfile = $ARGV[0])
{
   #################################################
   # Do following code block only if $curfile exists
   #################################################
   if (-e $curfile)
   {
      ###############################
      # stat structure into variables
      ###############################
      ($dev,$ino,$mode,$nlink,$uid,$gid,$rdev,$size,
      $atime,$mtime,$ctime,$blksize,$blocks)
      = stat("$curfile");
      ###############################
      # time structure into variables
      ###############################
      local($sec,$min,$hr,$day,$mon,$yr,$wday,@dntcare) = localtime($mtime);
      $yr = ($yr>=70) ? $yr+1900 : $yr+2000;
      $yr="$yr";
      $mon = (++$mon < 10) ? "0$mon" : "$mon";
      $day = ($day < 10) ? "0$day" : "$day";
      $hr  = ($hr < 10) ? "0$hr" : "$hr";
      $min = ($min < 10) ? "0$min" : "$min";
      $sec = ($sec < 10) ? "0$sec" : "$sec";
      ##################################################################
      # Rearrange in the YYYYMMDDhhmmss format and assign to $dte variable
      ##################################################################
      $dte = join(':',$yr,$mon,$day,$hr,$min,$sec);
      ######################################
      # Print modification date and filename
      ######################################
      print ("$dte $curfile\n");
      }
   ########################################
   # Shift to next position in command line
   ########################################
   shift (@ARGV);
}

# 4  
Old 11-09-2004
Quote:
If it is not showing the year in 'ls -l ' listing ... then it is abvious that the file is modified last time in the same year ;
That's true but what if I want the year of file creation in a variable
which I can use dynamically. I could have done
VARIABLE=`ls -l|awk '{print $6$7$8,2004}'`
for getting the dd:mm:yyyy format for the file but here the
value for 2004 is static and would be useless once the year changes.

For Jim's code I still have not tested it but looks promising to me.

Chill
enc
# 5  
Old 04-22-2005
Problem faced with files greater than 2 GB
Guys,
I used the following snippet as suggest above.

#!/usr/bin/perl -w
use strict;

foreach (@ARGV) {
my $mtime=(stat($_))[9];
my ($ss,$mm,$hh,$DD,$MM,$YY)=localtime($mtime);
printf "%04d %s\n",$YY+1900,$_;
}


Problem is when the file name passed as argument is greater then 2 Gb.

Any suggestions or a roundabout will be of great help. As most of my files I want are of huge size.

My OS is Sun Solaris 5.8.

thanks
# 6  
Old 04-22-2005
Try this script:
Code:
#! /usr/bin/ksh

typeset -Z2 d m
Jan=1 Feb=2 Mar=3 Apr=4 May=5 Jun=6 Jul=7 Aug=8 Sep=9 Oct=10 Nov=11 Dec=12
date "+%Y %m" | read year month

for i ; do
        line=$(ls -dl $i)
        line=${line##+([! ])+([ ])}
        line=${line##+([! ])+([ ])}
        line=${line##+([! ])+([ ])}
        line=${line##+([! ])+([ ])}
        line=${line##+([! ])+([ ])}
        set -A stamp $line
        d=${stamp[1]}
        eval m=\$${stamp[0]}
        y=${stamp[2]}
        ((${#y} != 4)) && ((y=year-(m>month)))
        echo $y $m $d $i
done
exit 0

# 7  
Old 04-23-2005
MySQL Thanks
Thanks!

It really works... trying to decipher the code Smilie
 
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