Finding executable files using ls and grep

Question

I have to write a script that finds all executable files in a directory. So I tried several ways to implement it and they actually work. But I wonder if there is a nicer way to do so.

So this was my first approach:

ls -Fla | grep \*$

This works fine, because the -F flag does the work for me and adds to each executable file an asterisk, but let's say I don't like the asterisk sign.

So this was the second approach:

ls -la | grep -E ^-.{2}x 

This too works fine, I want a dash as first character, then I'm not interested in the next two characters and the fourth character must be a x.

But there's a bit of ambiguity in the requirements, because I don't know whether I have to check for user, group or other executable permission. So this would work:

ls -la | grep -E ^-.{2}x\|^-.{5}x\|^-.{8}x

So I'm testing the fourth, seventh and tenth character to be a x.

Now my real question, is there a better solution using ls and grep with regex to say:

I want to grep only those files, having at least one x in the ten first characters of a line produced by ls -la

Answer 1

Do you need to use ls? You can use find to do the same:

find . -maxdepth 1 -perm -111 -type f

will return all executable files in the current directory. Remove the -maxdepth flag to traverse all child directories.

You could try this terribleness but it might match files that contain strings that look like permissions.

ls -lsa | grep -E "[d\-](([rw\-]{2})x){1,3}"