How do i print the path of excuatable in echo [duplicate]

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
– Kusalananda ♦, Commented Jan 18, 2021 at 11:35

raj · Accepted Answer · 2021-01-18 11:32:06Z

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)

Better with command -v exe. See unix.stackexchange.com/questions/85249 — Kusalananda
– Kusalananda ♦, Commented Jan 18, 2021 at 11:35

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