Test if executable exists in unix

Question

Problem

I may or may not have the executable thing in my $PATH.

How can I neatly check this in a ZSH script?

Existing Attempt

I just run the command, sending the output and errors or noman's land, then check the result code.

thing > /dev/null 2>&1
thing_installed=$?
if [ $thing_installed -eq 0 ]; then
    echo 'Thing Installed!'
fi

I feel like this could be done more neatly (one liner?).

Answer 1

In zsh, which behaves reasonably, so you can simply do

if which thing > /dev/null 2>&1; then
    echo installed
fi

or

which thing > dev/null 2>&1 && echo installed

Note that which is a shell builtin, and its behavior is not reasonable in all shells, so this behavior cannot be relied upon.

Answer 2

For zsh whence is the tool you are looking for. With parameter -p it can be forced to look only in $PATH and ignore functions, builtins and aliases with the same name.

This will either return the path to thing or an error message:

whence -cp thing

Answer 3

use which thing and check result code

Answer 4

You can check whether the file exists by this expresion:

[ -e thing ] && echo 'Thing Installed!'

The -e operator checks whether the file exists or not. If you want to do other file tests take a look to this page: http://tldp.org/LDP/abs/html/fto.html.

Hope this was helpful ;-)