Bash - Get specific argument using number in variable

Question

I want to access an argument from when the script was run, and I want to get the specific argument based on the variable. So I want to get the argument $2 but rather than typing two, I use the variable that contains a number

Something like this

FOO=2 echo $FOO

The problem is both variables and arguments both use $ so I don't know how to call an specific argument based on the number in the variable. I know I'm explaining this terribly, correct me if you can.

Eric Renouf · Accepted Answer · 2016-09-11 22:21:41Z

2

bash allows indirection with the ${!name} notation, where name is the name of a variable whose value is the name of the variable you want to access. So, in your example you could print the second positional parameter like

foo=2
printf '%s\n' "${!foo}"

or

printf 'The value at %s is %s\n' "$foo" "${!foo}"

which will print something like

The value at 2 is secondParameterValue

Some good reading on indirection can be found here

answered Sep 11, 2016 at 22:21

Eric Renouf

14.6k3 gold badges53 silver badges73 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Bash - Get specific argument using number in variable

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related