How to Find/Cut for Only The Filename from an Output of Ls -Lrt in Perl

How to find/cut for only the filename from an output of ls -lrt in Perl

You're making it harder for yourself by using -l. This will do what you want

print((`ls -brt`)[0]);

But it is generally better to avoid shelling out unless Perl can't provide what you need, and this can be done easily

print "$_\n" for (sort { -M $a <=> -M $b } glob "*")[0];

How to display only files from aws s3 ls command?

You can't do this with just the aws command, but you can easily pipe it to another command to strip out the portion you don't want. You also need to remove the --human-readable flag to get output easier to work with, and the --summarize flag to remove the summary data at the end.

Try this:

aws s3 ls s3://mybucket --recursive | awk '{print $4}'

Edit: to take spaces in filenames into account:

aws s3 ls s3://mybucket --recursive | awk '{$1=$2=$3=""; print $0}' | sed 's/^[ \t]*//'

Sort any random list of files by modification date

Your some_sorter_by_mtime should be for example:

xargs stat -f "%m %N" | sort -n | cut -f2-

the idea behind is:

• print out file modification time and the filename
• sort the output numerically (so by modification time)
• cut out the time field

so,

find / -type f -print | xargs stat -f "%m %N" | sort -n | cut -f2-

Is there a way to get path/to/file with ls + awk, sed, grep or similar tools?

Have you tried find ./delete -type f -ls (note the -ls -- that's the key :-) )? You should then be able to pipe the results through awk to filter out the fields you want.

Edit...
Another way you could do it is with a while loop, e.g.:

find ./delete -type f -ls | while read f1 blocks perms blocks owner group size mon day third file
do
    echo `basename $file` `dirname $file`
done

and add the bits you need into that.

Unix shell script to sort files depending on the 'date string' present in their file name

one way with awk and sort:

ls -1|awk -F'[_.]' '{s=gensub(/^([0-9]{4})(.*)/,"\\2\\1","g",$3);print s,$0}'|sort|awk '$0=$NF'

if we break it down:

ls -1|
awk -F'[_.]' '{s=gensub(/^([0-9]{4})(.*)/,"\\2\\1","g",$3);print s,$0}'|
sort|
awk '$0=$NF'

the ls -1 just example. I think you have your way to get the file list, one per line.

test a little bit:

kent$  echo "SSA_F13_12142012.request.done
SSA_F12_05122013.request.done
SSA_F14_01062013.request.done"|awk -F'[_.]' '{s=gensub(/^([0-9]{4})(.*)/,"\\2\\1","g",$3);print s,$0}'|
sort|
awk '$0=$NF'
SSA_F13_12142012.request.done
SSA_F14_01062013.request.done
SSA_F12_05122013.request.done

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