Is there a way to create a command line wrapper for a bash file?

Question

The title says it all. I would like to type "run" in the command line and it would execute /pathtoscript/script.sh.
Thank you!

Answer 1

Create a file named run with the contents:

#!/bin/sh
exec /pathtoscript/script.sh "$@"

Place the file in your path and set the executable bit.

What is "your path"?

At the command line, type:

echo $PATH

You will see a colon-separated list of directories. These are the directories that the shell searches when looking for a command to run. They are collectively called the path. You will want to place your new file run in any one of those directories.

Alternate approach for interactive work

If you just want run to work when you are working interactively, you can create an alias:

alias run=/pathtoscript/script.sh

If you want this alias saved permanently, put that line in the .bashrc in your home directory.

Answer 2

In *nix shells, the way this is done is by adding an executable file (or symbolic link to an executable file) to one of the directories listed in the PATH variable. From a hypothetical session:

$ which foo
which: no foo in (/usr/local/sbin:/usr/local/bin:/usr/bin)
$ foo
bash: foo: command not found
$ echo "$PATH"
/usr/local/sbin:/usr/local/bin:/usr/bin
$ sudo ln -s /path/to/my/script /usr/local/bin/foo
$ foo
[whatever your script prints]