UNIX for Dummies Questions & Answers

Hello,
I'm using ls -laR to print out a list of file and folders. I want to print only the permission, file size and file name. Also, excluding the '.' and '..'.

result from ls -laR:

total 6
drwxr-xr-x 8 user staff 512 Nov 28 16:17 .
drwxr-x--- 16 user staff 1024 Nov 28 16:17 ..
drwxr-xr-x 4 user staff 512 Nov 28 16:17 build
-rw-r--r-- 1 user staff 2672 Nov 28 16:17 build.xml

./build:
total 2
drwxr-xr-x 4 user staff 512 Nov 28 16:17 .
drwxr-xr-x 8 user staff 512 Nov 28 16:17 ..

ideal print result:

total 2
drwxr-xr-x 512 build
-rw-r--r-- 2672 build.xml

./build:
total 0

Any suggestion on how I can do this?

Code:

ls -laR | awk '{print $1,$5,$9}' | egrep  -v ' \.*$'

or just with awk:

Code:

ls -laR | awk '$9 !~ /^\.*$/{print $1,$5,$9}'

Can try to pipe it into this:

Code:

awk '/^[d\-lbcp].+/ {print $1,$2,$NF} /^[^d\-lbcp].+/ {print}'

Will work until filenames with blanks show up.

Thank you so much