How to auto source a shell script?

Question

How to make a shell script execute it's commands as source commands without having to use "source scriptname.sh" expression to launch the script? Basically what I want is to type ./scriptname.sh and it's commands to act as if source was used to execute the shell-script.

Answer 1

None of the Unix shells allow you to mark a script as one that should always be sourced the way you're hoping. What I would do is create an alias or procedure to do what you want and put it in my startup script. For example, in tcsh:

alias scriptname source \!\!

In bash, you could write it as a procedure:

scriptname( ) { source scriptname.sh $*; }

Answer 2

You can use a feature of bash that allows you to intercept user input and act on it, overriding any usual command execution.

Add the following to your bash startup script:

shopt -s extdebug
function auto_source_names {
    local CMD=$BASH_COMMAND
    if [[ -f "$CMD" ]] && [[ ! -x "$CMD" ]] ; then
        source "$CMD"
        return 1
    else
        return 0
    fi
}
trap 'auto_source_names' DEBUG

This causes the function to be called for every command executed. If it's the name of a file that isn't executable (otherwise we'd block legitimate ./foo.sh calls as well), it is sourced and no command executed. Be aware of files named ls or rm in the same directory. You could add additional safeguards, e.g. requiring .sh file extension or file telling you the name is a text file.

---

Example:

Create foo.sh:

FOO=bar

Then:

$ echo $FOO
$ foo.sh
$ echo $FOO
bar

Answer 3

The first line needs to contain the path to the interpreter:
#! /bin/bash

Then make the script file executable:
chmod +x scriptname.sh

Now you can run ./scriptname.sh to execute it.