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Listing files in a particular format

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Top Forums Shell Programming and Scripting Listing files in a particular format
# 8  
Old 08-23-2011
Hi,

I think I have clearly stated my requirements. But if still required the output to make the things more clear, I am posting it as below--

Right now my output is like below-

Code:
ll | awk ' !/^d/ { if(NR>1) {printf "%-29s\t\t%s\t%5s\t\t%s %s,%s\n", $9,$1,$5,$6,$7,$8} }'

a.out                                   -rw-------          0           Aug 5,06:55
dir_str.txt                             -rw-------         35           Aug 19,10:23
dont_make_test                          ----------          0           Aug 23,10:03
get_dir_stru.sh                         -rwx------        599           Aug 5,11:02
mm                                      -rwx------         85           Aug 23,09:34

But he output I want should be like this
Code:
a.out                                   600          0           Aug 5, 2011
dir_str.txt                             600         35           Aug 19, 2011
dont_make_test                          000          0           Aug 23, 2011
get_dir_stru.sh                         700        599           Aug 5, 2011
mm                                      700         85           Aug 23, 2011

And I am sorry Panyam that I mentioned "creation time", yes it's modification time.

Thanks.
# 9  
Old 08-23-2011
Hello,

I'm sorry but I was not much helpful. I searched few things in the forum and found this below useful.

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 in `ls -l|awk '!/^d/{print $NF}` ; do
        line=$(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 $i ${stamp[0]} $d","$y
done
exit 0

You might need to figure it out on how to add file permissions and other fields to the output as required.
Last edited by panyam; 08-23-2011 at 11:32 AM.. Reason: added to get only files
# 10  
Old 08-23-2011
Hi Panyam,

I forget to mention, my shell is 'sh', so will the code given by you will work on that shell?

Thanks,
Manu
# 11  
Old 08-23-2011
Try this and see if returns the desired output:

Code:
ll|awk -v "year=`date '+%Y'`" '$8!~":"{if(NR>1)printf"%-29s\t\t%s\t%5s\t\t%s %s,%s\n", $9,$1,$5,$6,$7,$8} {if(NR>1)printf"%-29s\t\t%s\t%5s\t\t%s %s,%s\n",  $9,$1,$5,$6,$7, year}'

This User Gave Thanks to dude2cool For This Post:
manubatham20
# 12  
Old 08-23-2011
Hehe,

This is not the exact solution. I am not that much efficient in Unix coding. But by seeing your code, I came to know, that you have searched : in $8 parameter and if it not found, then you printed $8 as it is (as it will be already a year format) and if its found, you replaced $8 with current year (that is WRONG, we need to compare the months as well in that case, if month of file modification is greater than the current month of the current year, then we need to do year=year-1 else year=year). How this will help you (eventually me Smilie ) to write a code for that.

I appreciate your efforts. But whatever you have done can be done by the code below as well--

Code:
ll | awk -v "yr=`date '+%Y'`" ' !/^d/ { if(NR>1) {printf "%-29s\t\t%s\t%5s\t\t%s %s,", $9,$1,$5,$6,$7}; if(index($8,":")==3) { print yr } else { print $8 
} }'

Thanks.
Last edited by manubatham20; 08-23-2011 at 01:52 PM..
# 13  
Old 08-23-2011
Can you past a line from ls that you think the awk code will not work against? i.e paste a line from your ls output where "month of file modification is greater than the current month of the current year"? Maybe I can modify the logic.
# 14  
Old 08-23-2011
Well, for that I need to change the date/time of the system, for that privilidges not granted to me.

Just set your system date to '29-Dec-2010', make a file, now again set your date to '2-Jan-2011', now execute your command. Let me know if your code work perfectly.

Regards,
Manu

---------- Post updated 08-24-11 at 01:16 AM ---------- Previous update was 08-23-11 at 10:28 PM ----------

Code:
#!/bin/sh
ll | awk -v "yr=`date '+%Y'`" -v "mnth=`date '+%m'`" 'BEGIN { split("Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec", MON, "|"); for(N=1; N<=12; N++) I[MON[N]]=N; } !/^d/ { if(NR>1) {printf "%-29s\t\t%s\t%5s\t\t%s %s,", $9,$1,$5,$6,$7}; if(index($8,":")==3) { if(I[$6]>mnth) { print yr-1 } else { print yr }} else { print $8 } }'

The above has fulfilled my purpose.

Thanks all for your time and support.
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