find folders with executable files

Question

I wrote a script to find all folders that contain executable files. I was first seeking a oneliner command but could find one. (I especially tried to use sort -k -u).

. The script works fine but my initial question remains: Is there a oneliner command to do that?

#! /bin/bash
find $1 -type d | while read Path
do
X=$(ls -l "$Path" | grep '^-rwx' | wc -l)
if ((X>0))
then
    echo $Path
fi
done

Possible duplicate: http://stackoverflow.com/questions/4458120/unix-find-search-for-executable-files There's your answer. — , Dec 08 '13 at 18:16
@HermanTorjussen: this question is about finding directories with executable files in them, a bit different from your proposed dup. — Mat, Dec 08 '13 at 18:17

score 3 · Accepted Answer · edited May 23 '17 at 11:57

3

Using find:

find $1 -type f -perm /111 -exec dirname {} \; | sort -u

This finds all files with permission 111 (i.e. rwx) but then we output only the directory name. To avoid duplicates, sort -u is used.

As pointed out by Paulo Almeida in the comments, this would also work:

find $1 -type f -perm /111 -printf "%h\n" | sort -u

edited May 23 '17 at 11:57

Community

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answered Dec 08 '13 at 18:20

pfnuesel

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The -perm +nnn form is deprecated, prefer /111 instead. – denarced Dec 08 '13 at 18:22
You can also use something like `-printf "%h\n"` to avoid having to call `dirname`. – Paulo Almeida Dec 08 '13 at 18:24
Thanks, edited my answer. I assume this solution would be faster since `dirname` calls a subprocess? – pfnuesel Dec 08 '13 at 18:29

find folders with executable files

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