How to make .bashrc call dependant on tmux program being installed?

Question

I have a section in my .bashrc that fires up the tmux program (terminal multiplexor).

However if the tmux program isn't installed (say I'm setting up a new computer) then having this in my .bashrc file stops any terminal window from successfully opening at all.

Of course installing tmux fixes this, but that isn't my problem.

How can I make this conditional so that if tmux isn't installed it doesn't crap out / give an error message?

Currently I have:

if [[ ! $TERM =~ screen ]]; then
  exec tmux
fi

I want something like:

if tmux; then
  if [[ ! $TERM =~ screen ]]; then
    exec tmux
  fi
fi

but that gives me

The program 'tmux' is currently not installed.  You can install it by typing:
sudo apt-get install tmux

Though it does at least give me a terminal prompt instead of closing the window! Also if/when tmux is installed this doesn't give any error / cause any issue when opening a new terminal window.

Answer 1

You can use the command type to see if the executable is present on your box:

if [ -n "$(type -P tmux)" ]; then

    ...tmux is installed...

else

    ...tmux isn't installed...

fi

I've often used this code snippet to do it:

$ [ -n $(type -P tmux) ] && echo "installed" || echo "not installed"
installed

I can fake it out using the alternative to -n (not empty string), -z (empty string).

$ [ -z $(type -P tmux) ] && echo "installed" || echo "not installed"
not installed

Answer 2

Or, if all you want to do is decide on which executable to fire off, you can do this:

executable=$(type -P tmux)
executable=${executable:-xterm} # For example
exec $executable

The expansion var=${var:-string} assigns the value "string" to var if and only if nothing has been assigned to var before.