grep not extracting desired string

Question

I'm working on the following question for my class (in Unix) and I'm running into some problems.

Using the grep command and regular expressions, create a pipe-based chain of commands that will list all files in the default directory that others can read or write.

This is the command I have so far and it will correctly match the right lines, but I only want to to print the name of the file that has the desired permissions.

ls -la | grep "^.......rw.*:\d\d\s(.*)$"

here is the result of just typing ls -la by the way which is what I'm "greping"

total 344
drwxr-xr-x  10 cameronpattisall  staff     340 Apr  7 11:21 .
drwxr-xr-x   7 cameronpattisall  staff     238 Apr  7 11:22 ..
-rw-r--r--   1 cameronpattisall  staff    6148 Apr  7 11:21 .DS_Store
-rw-r--r--   1 cameronpattisall  staff   11591 Apr  7 11:21 DifferencesToTagged.txt
-rw-r--r--   1 cameronpattisall  staff    4291 Apr  7 11:21 functions.py
-rw-r--r--   1 cameronpattisall  staff    6080 Apr  7 11:21 ls.txt
-rw-r--r--   1 cameronpattisall  staff    5511 Apr  7 11:21 prog4.py
-rw-r--r--   1 cameronpattisall  staff     345 Apr  7 11:21 shared.py
-rw-r--r--   1 cameronpattisall  staff    1832 Apr  7 11:21 subversion-log.txt
-rw-r--r--   1 cameronpattisall  staff  122295 Apr  7 11:21 testcases.txt

Answer 1

I guess the solution is to use:

grep -E "^-.{6}(r|.w)"

The structure of the first column is "-/d" (file/dir) + xxx (owner) + xxx (group) + xxx (others).

So you want to match those starting with - (file) and containing in the "gorup" part either r.. or .w..

To print just the file names, then you can pipe to grep -o '[^ ]*$'. This will print the last block of text. Note this is terrible fragile, since in a file name with spaces it will just print the last one.

Test

I stored your ls -la in the file a:

$ grep -E "^-.{6}(r|.w)" a | grep -o '[^ ]*$'
.DS_Store
DifferencesToTagged.txt
functions.py
ls.txt
prog4.py
shared.py
subversion-log.txt
testcases.txt