find command: how to find all such files with a mask?

I'm assuming your criterion for avoiding a file/directory is that it has any permission bits set for group or other. You can use either of these constructs to avoid matching such files:

find <path> ! ( <permissions_mask_set> ) -exec ls -ld {} ; find <path> ( <permissions_mask_set> ) -o -exec ls -ld {} ; 

Taking the first one of these because that matches your own example, it would become:

find . \! \( -perm -001 -o -perm -002 -o -perm -004 -o -perm -010 -o -perm -020 -o -perm -040 \) -exec ls -ld {} \; 

For systems with GNU find the permissions mask can be much simplified:

find . \! -perm /077 -exec ls -ld {} \; 

The reason your own attempt is showing everything, including your excluded files, is because one of the values passed to the -exec is a command that looks (partially) like this

ls -l . 

If you run that at the command line you'll get all the (non-"dot") files/directories in the current directory listed - including any you want excluded. The key is either to use the -d flag to avoid listing the contents of directory, or to qualify the find command either looking only for files or excluding directories.

On Solaris (which BTW is no longer BSD-based), you should be able to use GNU find where -perm accepts a /bitset which evaluates to true if any of the file permission bits are in bitset.

/usr/gnu/bin/find . ! -perm /77 -ls 

Or

/usr/gnu/bin/find . ! -perm /go=rwx -ls 

Would return the files where none of the ---rwxrwx bits are set.

Or you could use zsh which should be installed by default on Solaris as long as you're not using a stripped down flavour:

ls -ld -- **/*(Df00) 

Or:

ls -ld -- **/*(Df-77) 

Or:

ls -ld -- **/*(Df[go=]) 

Or:

ls -ld -- **/*(D^AIERWX) 

From another shell:

zsh -c 'ls -ld -- **/*(Df[go=])' 

For instance.