Use ls command without showing permissions

Howard_L · #1 19-Oct-2008, 09:54

Hi, file names get long and it's nice to reduce the left side of ls -l printout. 'man ls' shows me I can:

Code:

ls -go   -no group, no owner.
ls -got  -no group, no owner, sort by time (ascending)
ls -gotr -no group, no owner, sort by time descending (r= reverse sort)
eg:
$ ls -gotr
total 488
-rwxr-xr-x 1  5979 Sep  8 22:32 panel_x-1.0
-rw-r--r-- 1  1066 Sep  8 22:40 panel_x-1.0.c
-rwxr-xr-x 1  8557 Sep  8 22:42 panel_x-2.0
-rw-r--r-- 1  3545 Sep  8 22:55 panel_x-2.0.c
-rw-r--r-- 1  3801 Sep  9 08:51 books_curses_cpp-1.0.cpp.bak

But I'd like to dump the permissions too.
Anybody know how?

Howard_L · #2 20-Oct-2008, 00:03

I have been enlightened to two methods so far:

ls -latr | awk '{printf("%7s %s %2s %5s %s\\n",\$5,\$6,\$7,\$8,\$9)}'";
and
ls -got | cut -b 1,13-128

and I made shell script to use it as 'myls' which works for my pwd:

Code:

#!/bin/sh   
echo "  ~/bin/myls  custom ls command   20081028"; 
echo "man ls , awk , or cut for more info, but:"
echo "  -g nogroup, -o noowner, -t sort by time(ascending), -r reverse order";
echo "Was using:  ls -latr | awk '{printf("%7s  %s %2s %5s  %s\\n",\$5,\$6,\$7,\$8,\$9)}'";
echo "Now using:  ls -got | cut -b 1,13-128 " ;
echo "Listing of: $(pwd):"
ls -got | cut -b 1,13-128

TurboPT · #3 07-Nov-2008, 21:07

Hey Howard,

Here's another way to "reduce the left": (similar to one of your others -- but no sorting/reversing in this snippet)

Code:

[root@localhost ~]# ll    // <<--- aliased as ls -l on my system.
total 128
-rw------- 1 root root  1092 Jul  7  2006 anaconda-ks.cfg
drwxr-xr-x 4 root root  4096 Oct 27 23:12 Desktop
-rw-r--r-- 1 root root 24050 Jan 13  2008 EcllipseConfig
-rw-r--r-- 1 root root 44248 Jul  7  2006 install.log
-rw-r--r-- 1 root root  5242 Jul  7  2006 install.log.syslog
-rw------- 1 root root  1645 Nov  6 21:58 mbox
-rw-r--r-- 1 root root   706 Sep 22 23:24 test.php
drwxr-xr-x 5 root root  4096 Feb  3  2007 workspace

[root@localhost ~]# ll | awk 'BEGIN{OFS="\t"}{print $5,$6,$7,$8,$9}'

1092    Jul     7       2006    anaconda-ks.cfg
4096    Oct     27      23:12   Desktop
24050   Jan     13      2008    EcllipseConfig
44248   Jul     7       2006    install.log
5242    Jul     7       2006    install.log.syslog
1645    Nov     6       21:58   mbox
706     Sep     22      23:24   test.php
4096    Feb     3       2007    workspace

OFS in the BEGIN block of the awk command implies: (O)utput (F)ield (S)eparator.

Howard_L · #4 09-Nov-2008, 08:35

linux tools like awk are so cool...
You know these are available 'standalone' for the msdos console as well.
But how to get the first character of $1.... for directory (or file)
That's why I've ended up using this so far:

Code:

-----  ~/bin/myls  -----
#!/bin/sh
echo "Listing of: $(pwd):"
ls -got | cut -b 1,13-128
-----  out:  -----
Listing of: /mnt/e/ccpl:
t
d 4096 Nov  8 09:24 cgi-bin
d 4096 Nov  5 10:47 sandbox
-  518 Nov  4 19:11 index.html

davekw7x · #5 09-Nov-2008, 11:19

Quote:

Originally Posted by Howard_L

...But how to get the first character of $1...

First of all: If you have a way that works for you: Rejoice.

On the other hand, if you want to investigate awk a little more, you could check out awk's printf. It's a lot like printf() for C

So, you could try something like:

Code:

ls -l | awk '{printf("%c  %10s %3s %2s %5s %s\n", $1,$5,$6,$7,$8,$9)}'

See? The "%c" for the first field just prints the first character of the field.

Adjust the %s field widths to give it the look that you like.

Left as an exercise for the student: The first line of my "ls -l" is "total xxxxx", which results in a line with the single character 't' as the first line of the output when I run that awk script.

Your assignment:

Either
1. Get rid of that line in the awk output

or, better

2. Print the line exactly as ls -l gives it to you.

Regards,

Dave

Footnote: What if some of your file names had spaces as part of their names?

"I pity the fool..."
---Mr. T

TurboPT · #6 09-Nov-2008, 18:13

Quote:

Originally Posted by Howard_L

But how to get the first character of $1.... for directory (or file)

...alternatively, the -F option with ls command will append a slash to the end of a name that is a directory. See this:

Code:

[root@localhost ~]# ls -lF | awk 'BEGIN{OFS="\t"}{print $5,$6,$7,$8,$9}'

1092    Jul     7       2006    anaconda-ks.cfg
4096    Oct     27      23:12   Desktop/
24050   Jan     13      2008    EcllipseConfig
44248   Jul     7       2006    install.log
5242    Jul     7       2006    install.log.syslog
1645    Nov     6       21:58   mbox
706     Sep     22      23:24   test.php
4096    Feb     3       2007    workspace/

EDIT:
As Dave pointed out though, this example (and the previous post) falls short if a filename contains spaces.

Howard_L · #7 15-Nov-2008, 19:24

I had left this tab open on my browser to make a point of getting back to it.
yes I have not needed to reboot for a week or so in fact:

Code:

$ uptime
 21:20:13 up 30 days,  9:16,  1 user,  load average: 0.27, 0.18, 0.09

That's with all four virtual windows going with two or three xterms open in each plus etc etc...
(as opposed to the XP box which slows to a crawl after a couple of days)
...anyhow man awk is overwhelming, I see this:vectorsite.net/tsawk.html
and am quickly clued to do this:

Code:

$ echo "  pwd: `pwd`" && ls -l | awk  \!'/total/ {printf("%c  %10s %3s %2s %5s %s\n", $1,$5,$6,$7,$8,$9)}'
  pwd: /var/www/cgi-bin
-         638 Nov 13 22:26 hello-1.pl
-         764 Nov 14 10:38 hello-2.pl
-        1474 Nov 14 15:52 hello.pl

cool dude...
%10s dave? you must have some big files on there!

davekw7x · #8 15-Nov-2008, 21:17

Quote:

Originally Posted by Howard_L

Code:

$ echo "  pwd: `pwd`" && ls -l | awk  \!'/total/ {printf("%c  %10s %3s %2s %5s %s\n", $1,$5,$6,$7,$8,$9)}'

But what if one of your files had the string "total" as part of its name?

If you don't want it to print the first line (record) you could make the awk part look like

Code:

awk 'NR>1 {printf("%c  %10s %3s %2s %5s %s\n", $1,$5,$6,$7,$8,$9)}'

Regards,

Dave

Howard_L · #9 15-Nov-2008, 23:50

Right dave, I'd thought about that and thought a $1 != kindof thing might work and find that this does:

Code:

 $ ls -la | awk '{if(! match($1 , /^total$/)) printf("%c  %10s %3s %2s %5s %s\n", $1,$5,$6,$7,$8,$9)}'
d        4096 Nov 16 01:49 .
d        4096 Nov 16 01:05 ..
-        1509 Nov 16 00:57 awk_notes
-         450 Nov 15 20:34 coins.txt
d        4096 Nov 16 01:49 total_mess

...but your NR method is much cleaner... The man pages are becoming more clear to me now.
Now I'm looking at those dot directories . hmmm $9 something...

davekw7x · #10 16-Nov-2008, 08:33

Quote:

Originally Posted by Howard_L

...
Now I'm looking at those dot directories . hmmm $9 something...

This is where your C programming experience can be useful. Maybe the conditional can be something like

Code:

{if ((NR > 1) && ($9 != "." ) && ($9 != "..")) do_some_stuff}

IWFM---YMMV

Regards,

Dave

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Use ls command without showing permissions

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