Bashscripts.org

I want to make a simple script to list all
files that are executable, many of them
are not *.sh, and are extensionless

any help appreciated,

Ted.

find has an -executable test.

you could also do a recursive grep to look for shebang pattern.

like this
find . -perm 755

but that's not what I want because then I get all files with those permissions.. maybe I really mean extensionless files.. files without the '.'

may be:

If I'm not completely mistaken it could be done like

(wich I believe is a good reason for the -executable option)
and

(regexes are overrated)

I don't get why you guys insist on that -executable option.. that doesn't exist or does it?

find . -type f -executable -regex "[^.]*"
find: -executable: unknown option

find . ! -name '*.*'
Also I'm getting all the .svn output with your last suggestions which is close to what I want, but not quite.. how would I omit the .svn dir from this output.

Ted.

I don't know about the -executable option, at least my manpage doesn't list it- Might be a newer version or a completely different find, I simply don't know.

For the .svn directory exclusion: Simply use -prune, it's made for this kind of job. http://mywiki.wooledge.org/UsingFind#A-prune

comes with -executable test option.

if your find does not have -executable then the following would probably be ok, if a little wordy;

Hi DarthWavy,

unfortunately, this won't work, because bash doesn't know {}:but still the -perm /mode option is shorter

I'm a little confused as this is not the line I submitted as an option. If you're going to imply that I said something I didn't then please make it uber so that I look good