Basic Scripting help
 

8. Create a local variable within your shell by typing the command VAR_ONE=variable1. Create an environment variable by typing the command export VAR_TWO=variable2.

Run the command echo "$VAR_ONE, $VAR_TWO", what output appears?
Answer: Obviously when I run this I get variable1,variable2


9. Create a shell script called echo.bash that contains the following lines:

VAR_3=three
export VAR_4=four
echo "$VAR_ONE, $VAR_TWO, $VAR_3, $VAR_4"

Give the script execute permissions and execute it by running ./echo.bash. Record the output of the command. (Note, this command requires the variables VAR_ONE and VAR_TWO to exist from the previous question.)
Answer: , variable2, three, four (why the comma as the first variable? is that because variable1 was not exported?

10. Run the command echo "$VAR_3, $VAR_4" do the variables VAR_3 and VAR_4 exist in your current shell?
Answer: , (Again, why just a comma? because neither was exported to the child process?)

Anyone that can explain these a little, woule be awesome!

Dru*;

At first glance, I thought you had posted a cut and paste homework assignment without comment. I'll highlight what appear to be your comments, but please check to see that I did it correctly.

Yeah, thats correct. Whats in blue is what I get when running the commands I am told to.

Try the commands with no commas and see what happens!

are you talking about when I echo it?

Yes
Not quite. For your child process idea try this:

Put echo "$VAR_3, $VAR_4" in another script called child.bash.
Now put the following line at the end of echo.bash - ./child.bash

This should help to understand the child process getting variables

Whereever the question applies!! If you are running a command with commas, then try it without commas.

Doing this gives me the same output minus all the commas