How to run every script in a directory except itself?

Question

I have a folder full of *.command files on my OS X workstation.

(For those that don't know, *.command files are just shell scripts that launch and run in a dedicated Terminal window).

I've dragged this folder onto my Dock to use a "stack" so I can access and launch these scripts conveniently via a couple clicks.

I want to add a new "run-all.command" script to the stack that runs all the *.command files in the same stack with the obvious exception of itself.

My Bash chops are too rusty to recall how you get a list of the *.command files, iterate them, skip the file that's running, and execute each (in this case I'd be using the "open" command so each *.command opens in its own dedicated Terminal window).

Can somebody please help me out?

Answer 1

How about something like this:

#! /bin/bash                                                                    

for x in ./*
do
        if [ "$x" != "$0" ]
        then
            open $x
        fi
done

where $0 automatically holds the name of the script that's running

Answer 2

Using @bbg's original script as a starting point and incorporating the comments from @Jefromi and @Dennis Williamson, and working out some more directory prefix issues, I arrived at this working version:

#!/bin/bash
for x in "$(dirname $0)"/*.command
do
  if [ "$(basename $x)" != "$(basename $0)" ]
  then
    open "$x"
  fi
done