How can I check if a command exists in a shell script?

Try using type:

type foobar

For example:

$ type ls
ls is aliased to `ls --color=auto'

$ type foobar
-bash: type: foobar: not found

This is preferable to which for a few reasons:

1. The default which implementations only support the -a option that shows all options, so you have to find an alternative version to support aliases

2. type will tell you exactly what you are looking at (be it a Bash function or an alias or a proper binary).

3. type doesn't require a subprocess

4. type cannot be masked by a binary (for example, on a Linux box, if you create a program called which which appears in path before the real which, things hit the fan. type, on the other hand, is a shell built-in (yes, a subordinate inadvertently did this once).

Five ways, 4 for bash and 1 addition for zsh:

• type foobar &> /dev/null
• hash foobar &> /dev/null
• command -v foobar &> /dev/null
• which foobar &> /dev/null
• (( $+commands[foobar] )) (zsh only)

You can put any of them to your if clause. According to my tests (https://www.topbug.net/blog/2016/10/11/speed-test-check-the-existence-of-a-command-in-bash-and-zsh/), the 1st and 3rd method are recommended in bash and the 5th method is recommended in zsh in terms of speed.

In general, that depends on your shell, but if you use bash, zsh, ksh or sh (as provided by dash), the following should work:

if ! type "$foobar_command_name" > /dev/null; then
  # install foobar here
fi

For a real installation script, you'd probably want to be sure that type doesn't return successfully in the case when there is an alias foobar. In bash you could do something like this:

if ! foobar_loc="$(type -p "$foobar_command_name")" || [[ -z $foobar_loc ]]; then
  # install foobar here
fi

The question doesn't specify a shell, so for those using fish (friendly interactive shell):

if command -v foo > /dev/null
  echo exists
else
  echo does not exist
end

For basic POSIX compatibility, we use the -v flag which is an alias for --search or -s.