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How can I count executable binaries on my Linux box?

I've been looking at your script to count executable binaries in the book Wicked Cool Shell Scripts and I think that there's a problem in the script. Specifically, what happens if I have a directory in my PATH more than once?

Dave's Answer:

Great point. The script you're talking about is called how-many-commands.sh:

#!/bin/sh

# how many commands: a simple script to count how many executable commands
# are in your current PATH.

myPATH="$(echo $PATH | sed -e 's/ /~~/g' -e 's/:/ /g')"

count=0 ; nonex=0

for dirname in $myPATH ; do
directory="$(echo $dirname | sed 's/~~/ /g')"
if [ -d "$directory" ] ; then
for command in $(ls "$directory") ; do
if [ -x "$directory/$command" ] ; then
count="$(( $count + 1 ))"
else
nonex="$(( $nonex + 1 ))"
fi
done
fi
done

echo "$count commands, and $nonex entires that weren't marked executable"

exit 0

The problem is indeed stuck in that script, because when I break down the PATH, I manage the problem of having spaces in the directories neatly (I convert them to '--' then convert them back after the "for" loop) but what if there's more than one occurrence of a directory in the path?

Turns out that's pretty easily done by having a bit more code in the myPATH statement:

myPATH="$(echo $PATH | sed -e 's/ /~~/g' -e 's/:/ /g' | sort | uniq)"

Can you see what I added? Not too difficult, but I'm using the super-helpful "uniq" command to remove duplicates, but it doesn't work until you ensure that your path entries are in order, hence the use of the word "sort").

Now you'll have an accurate executable / non-executable file count for your system and your set PATH value.

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Categorized: Shell Script Programming (Article 6942)
Tagged: bash, linux, shell scripting, shell scripts
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