select fields with sed

Question

Hey,

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

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

Answer 2

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 3

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 4

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

Answer 5

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