How do i print the path of excuatable in echo

Question

SO i have a Bash Script like this and goal is to determine the path of an executable file and i want to print it, here is what i'm doing

#!/bin/bash
exepath=which exe
echo "$exepath"

Now this instead of printing path its starting the exe file in my system, how do i print the path stored in exepath variable.

Is outputting the pathname all you want to do in your script? In that case, the variable is not needed and you could do it with just command -v exe. — Kusalananda, Jan 18 '21 at 11:35

score 2 · Answer 1 · answered Jan 18 '21 at 11:32

2

Instead of

exepath=which exe

(this command just runs exe, previously setting the environment variable exepath to literal value which)

you should use

exepath=`which exe`

or

exepath=$(which exe)

answered Jan 18 '21 at 11:32

raj

1

Better with command -v exe. See https://unix.stackexchange.com/questions/85249 – Kusalananda Jan 18 '21 at 11:35

1 Answers1