Exporting just a single file

Question

To make scripts accessible from anywhere in the terminal, one can use the following sequence of commands:

PATH=$PATH:some_path_here
export PATH

Which includes all files at the path sent in alongside all of its subfolders. Which means from then on just typing the name of a script stored within this folder structure will execute it.

How do I export only a single file this way instead of an entire folder alongside all its subdirectories?

Answer 1

Seems alias is what you need. You can specify any word to alias and set what will be execute when you execute your word.

alias newcommand='/path/to/your/command'

You can put it in you ~/.shellrc (where shell is your shell, so for bash it will be ~/.bashrc) and it will be available after start.

For more information look here.

Answer 2

Create a symbolic link to your file, and place it in a suitable PATH'd directory, eg $HOME/bin or whatever.

ln -s "/path/to/my-file"  "$HOME/bin/my-file" 

Answer 3

To explain the background for Rush's answer, export does not do anything with the executables in $PATH. export PATH simply means "I want the PATH variable to be inherited by all child processes." And what happens when you type myprogram on a prompt and press Enter is a series of lookups within the shell.

When a command is specified in BASH without a pathname (e.g. myprogram, or ls), and it isn't an alias, function, builtin or keyword, BASH searches through the directories in PATH, in order from left to right, to see whether they contain an executable of the name you typed.

$PATH, in other words, is a list of directories, which is why it can't be used for single commands (except implicitly, by putting the executable in a separate directory). You can't promote a command to a builtin or keyword, but you can create a function or alias. See the previous link for a simple explanation of their differences.