Find executable files recursively

Question

I have got the directory called Test and a few directories inside it. Both Test and the directories inside it have executable files. I'd like to print them with ls. I'd use this command.

ls -l `find Test/ -perm /u=x,g=x,o=x -type f`

Is this a good/right/quick command or not?

My solution is:

find Test/ -executable -type f -exec ls -l {} \;

and got the same result as warl0ck and pradeepchhetri offered.

Answer 1

Not really, you can integrate the ls command with find,

find Test/ -type f -perm /u=x,g=x,o=x -exec ls -l {} \;

UPDATE

Actually -executable is not an equivalent of -perm /u=x,g=x,o=x. You might have files that is executable only by the group or others, which will not be displayed.

So, depends on your purpose, if you want files executable only by you, that's okay to use -executable.

Answer 2

There is no need to use -exec, as find comes with a -ls flag.

$ find Test/ -perm /u=x,g=x,o=x -type f -ls

From the man page:

-ls True; list current file in ls -dils format on standard output. The block counts are of 1K blocks, unless the environment variable POSIXLY_CORRECT is set, in which case 512-byte blocks are used. See the UNUSUAL FILENAMES section for information about how unusual characters in filenames are handled.

Answer 3

You should use -exec argument of find command.

$ find Test/ -perm /u=x,g=x,o=x -type f -exec ls -l {} \;

Answer 4

Find files that are executable by the current login user

[sudo] find /path -type f -executable

or to find if executable bit sets for at least one of the Owner/Group/Other:

[sudo] find /path -type f -perm /111

or executable only to owner:

[sudo] find /path -type f -perm /100

or

[sudo] find /path -type f -perm -100