Counting number of executable files in every variable inside PATH

Question

I want to write a script that would count number of executable files in every folder in variable PATH. My code:

#!/bin/bash

IFS=":"
for directory in $PATH; do
        files=0
        ls -l $directory | while read rights x name group siz m d h name; do
                if [ `echo $rights | cut -c1` = "-" ]; then
                        files=$((${files}+1))
                fi
        done
        echo "Directory ${directory} contains ${files} executable files"
done

I want the echo to process after the end of the while loop and before the start of next for loop, but it always print out number of files = 0. The counting inside the if condition works.

[Why *not* parse `ls`?](http://unix.stackexchange.com/questions/128985/why-not-parse-ls) — Cyrus, Apr 22 '18 at 14:01
This can be one line of php, but as usual i will be downvoted by noobs. php is better than bash itself for bash scripts... in every aspects. — NVRM, Apr 22 '18 at 14:55

score 0 · Accepted Answer · answered Apr 22 '18 at 13:30

0

Have you considered using the find command? See this link. On GNU-based versions of find, something like the following should replace your ls call and while loop.

find $directory -type f -executable -print | wc -l

If you are worried about traversing subdirectories, you can use the maxdepth flag, i.e.

find $directory -maxdepth 1 -type f -executable -print | wc -l

For BSD-versions, again, see the link.

answered Apr 22 '18 at 13:30

djkern

452
3
11

Piping the output of `find -print` to `wc` is no better than using the output of `ls`. – chepner Apr 22 '18 at 14:40

Counting number of executable files in every variable inside PATH

1 Answers1