Bash script: Can't set a variable using which

Question

In my terminal,

prog="cat"
name=$(which $prog)
echo $name

prints /bin/cat

But in my script:

pro="$1"
prog=$(which $pro)
echo "pro is $pro"
echo "prog is "$prog""

running scriptname cat prints

pro is cat
prog is

How do I make which work? it should print prog is /bin/cat

Can't reproduce. Does which succeed? What happens if you set -ex? — Biffen
– Biffen, Commented Jan 25, 2016 at 20:58
Sorry, it turns out it just wasn't working for my own scripts, not bash programs. — user
– user, Commented Jan 25, 2016 at 21:03

Rany Albeg Wein · Accepted Answer · 2016-01-25 21:03:03Z

1

which(1) is an external program used to search PATH for an executable. It behaves differently on different systems and you can't rely on a useful exit code; use (from most to least portable) command -v or type -P (to find the path) or hash (to check) instead.

try printf '%s\n' "$PATH" inside your script as well as outside of it. maybe the command you're looking for is not in the PATH used in the script?

That is almost certainly the cause.

answered Jan 25, 2016 at 21:03

Rany Albeg Wein

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2 Comments

user1934428 Over a year ago

Since which can be executed, it means which is in the PATH. It would be a very odd installation if cat would not be in the same directory as which, so it should be in the PATH too.

Rany Albeg Wein Over a year ago

@user1934428 Specifically speaking about cat and alike, you're correct. Thank you for leaving a comment and improving my answer!

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