"ls" command output

Question

I'm running ls -l which brings the following output:

-rw-r----- 1 webadm webgrp 168 Jul 19 16:00 SYNCHRO_20120719_1600.csv

Is there a way to have just get the last two fields? i.e:

16:00 SYNCHRO_20120719_1600.csv

If we're going to mark this one off-topic, shouldn't it at least be migrated to somewhere appropriate (ie. superuser)? — Charles Duffy, Jul 19 '12 at 15:13

Levon · Answer 1 · 2012-07-19T14:51:28.853

1

try this:

 ls -l | awk '{print $(NF-1), " ", $NF}'

It will display the next-to-last and last fields on the line, which is the time and the name of the directory entry and is not dependent on a specific number for a field.

May create problems with names that contains blanks however.

See this page regarding use of 'ls' for processing, provided by @CharlesDuffy which somewhat undermines the approach above :-) .. so the quick solution presented ought to be viewed in that context.

edited Jul 19 '12 at 14:51

answered Jul 19 '12 at 14:17

Levon

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Not just names with blank spaces -- how `ls` munges names in general (including ones with other forms of non-printable characters) isn't well-defined across implementations. Much better to use shell globbing to get the names, and something else (which _could_ be ls) to read their times. See http://mywiki.wooledge.org/ParsingLs – Charles Duffy Jul 19 '12 at 14:28
@CharlesDuffy Thanks for the feedback, useful information and so noted. I agree, this is an awkward (and quite possibly "iffy") solution. – Levon Jul 19 '12 at 14:31

score 1 · Answer 2 · answered Jul 19 '12 at 14:19

1

This should do what you want:

ls -l | awk '{print $8" "$9}'

answered Jul 19 '12 at 14:19

Costi Ciudatu

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This won't work as-is with filenames with spaces. – Charles Duffy Jul 19 '12 at 14:22
For my linux account I had to use fields $7 and $8 – Levon Jul 19 '12 at 14:26
This not at all meant to be a general solution. It's just the simplest command I could find that is able to handle the file format as shown in the question. – Costi Ciudatu Jul 19 '12 at 14:34

score 1 · Answer 3 · answered Jul 19 '12 at 14:27

On a system with GNU date:

for f in *; do
    printf '%s %s\n' "$(date +%H:%M -r "$f")" "$f"
done

This will work correctly with filenames with spaces -- or even filenames with newlines, though it won't be possible to distinguish between filename-embedded newlines and record-separated ones. (Use null delimiters if this is important).

"ls" command output

3 Answers3