select fields with sed - bash

Question

Using only grep and sed, is there a way I can tranform the output of ls -l * into this :
-rw-r--r-- agenda.txt
-rw-r--r-- annuaire.txt
Thanks!

Answer 1

seeing that you have already got your "answer", here's one of the simpler solution
ls -l | tr -s " "| cut -d" " -f1,8-
#OP, sed is "powerful", but sometimes, simplicity is more powerful.
Side note: Don't parse file names like that.

Answer 2

ls -l | sed 's/[ ]+//g' | sed 's/ [0-9].*:.[0-9]/ /g'

Answer 3

ls -altrh| sed -E 's/ +.+ / / g'
Or you can go with ssed which supports Perl Regular Expressions.
I solved your problem using the ssed program you can install it in any Posix system, ssed stands for super sed.
so i did a ls -latrh in my home directory.
telsa:~ mahmoh$ ls -altrh
total 136
drwxr-xr-x 5 root admin 170B Jun 24 00:27 ../
drwx------+ 4 mahmoh staff 136B Jun 24 00:27 Pictures/
drwx------+ 3 mahmoh staff 102B Jun 24 00:27 Music/
drwx------+ 3 mahmoh staff 102B Jun 24 00:27 Movies/
drwx------+ 3 mahmoh staff 102B Jun 24 00:27 Desktop/
-rw------- 1 mahmoh staff 3B Jun 24 00:27 .CFUserTextEncoding
drwxr-xr-x+ 5 mahmoh staff 170B Jun 24 00:27 Public/
drwx------+ 5 mahmoh staff 170B Jun 24 02:19 Documents/
-rw-r--r--# 1 mahmoh staff 15K Jun 24 02:19 .DS_Store
drwx------# 36 mahmoh staff 1.2K Jun 24 14:48 Library/
-rw-r--r-- 1 mahmoh staff 279B Jun 24 15:27 .profile~
-rw-r--r--# 1 mahmoh staff 14K Jun 24 15:29 .vimrc
-rw-r--r-- 1 mahmoh staff 279B Jun 24 15:30 .profile
drwx------ 2 mahmoh staff 68B Jun 24 15:46 .Trash/
drwxr-xr-x 3 mahmoh staff 102B Jun 24 20:26 .mplayer/
-rw------- 1 mahmoh staff 3.5K Jun 24 22:11 .bash_history
-rw------- 1 mahmoh staff 42B Jun 24 23:25 .lesshst
-rw-r--r-- 1 mahmoh staff 3.6K Jun 24 23:39 temp
-rw-r--r-- 1 mahmoh staff 3.3K Jun 24 23:43 rtorrent.rc~
drwxr-xr-x 5 mahmoh staff 170B Jun 24 23:52 torrents/
-rw-r--r-- 1 mahmoh staff 3.3K Jun 24 23:56 .rtorrent.rc~
-rw------- 1 mahmoh staff 3.7K Jun 24 23:56 .viminfo
-rw-r--r-- 1 mahmoh staff 3.3K Jun 24 23:56 .rtorrent.rc
drwxr-xr-x+ 25 mahmoh staff 850B Jun 24 23:56 ./
drwx------+ 10 mahmoh staff 340B Jun 24 23:58 Downloads/
Now watch.
telsa:~ mahmoh$ ls -altrh| ssed -R -e 's/ +.+ / / g'
total 136
drwxr-xr-x ../
drwx------+ Pictures/
drwx------+ Music/
drwx------+ Movies/
drwx------+ Desktop/
-rw------- .CFUserTextEncoding
drwxr-xr-x+ Public/
drwx------+ Documents/
-rw-r--r--# .DS_Store
drwx------# Library/
-rw-r--r-- .profile~
-rw-r--r--# .vimrc
-rw-r--r-- .profile
drwx------ .Trash/
drwxr-xr-x .mplayer/
-rw------- .bash_history
-rw------- .lesshst
-rw-r--r-- temp
-rw-r--r-- rtorrent.rc~
drwxr-xr-x torrents/
-rw-r--r-- .rtorrent.rc~
-rw------- .viminfo
-rw-r--r-- .rtorrent.rc
drwxr-xr-x+ ./
drwx------+ Downloads/

Answer 4

ls -l | sed 's/^\([^\t ]\+\)\(.*:.[^ \t]\+\)\(.\+\)/\1 \3/'

Answer 5

Here is a working command. The slightly tricky thing is that ls -l will print the year for files that are older than some time (6 months) and hh:mm for newer files.
ls -l | sed 's/ .*[0-9]* .*[A-Z][a-z][a-z] [ 0-9][0-9] \{1,2\}[0-9][0-9]:*[0-9][0-9] / /'
For the following example
drwxr-xr-x 39 root root 1024 Feb 19 08:58 /
the starting .* will match 39 root root 1024 and then the rest of the regular expression matches month name (so you might restrict a-z to fewer characters) followed by year or hh:mm.

Answer 6

why not use awk instead of sed? awk is built for stuff like this.
see this manual page for more about fields in awk.

Answer 7

Like this?
ls -l | sed 's/ [0-9].*:.[0-9] / /' | less
Transforms
-rw-r--r-- 1 tomislav tomislav 609 2009-11-26 10:32 Test.class
-rw-r--r-- 1 tomislav tomislav 46 2009-12-14 12:16 test.groovy
into
-rw-r--r-- Test.class
-rw-r--r-- test.groovy