Variable substitution in Shell script

Question

I have declared one variable IS_abc=false, on basis of certain condition I am changing value to IS_abc=true

IS_abc=false
declare -a my_arr
my_arr = ('abc' 'pqr' 'xyz')
....
.... // some operation
IS_abc=true
for i in "${my_arr[@]}"
do
    //here i want to access value of $IS_abc  as true
    //how to do this
done

I have tried accessing using $IS_'$i' , but it raising error as invalid substitution

Tell me if I am doing anything wrong here?

anubhava · Accepted Answer · 2017-06-22 07:33:14Z

4

You can use indirect var reference:

my_arr=('abc' 'pqr' 'xyz')
IS_abc=true

var="IS_${my_arr[0]}"
echo "${!var}"

Output:

true

answered Jun 22, 2017 at 7:33

anubhava

790k67 gold badges603 silver badges671 bronze badges

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

2 Comments

papkass Over a year ago

What does the ! do in "${!var}"?

anubhava Over a year ago

That is called indirect expansion. This expands to the name of the variable referenced by parameter. So it prints value of IS_abc variable.

papkass · Accepted Answer · 2017-09-11 11:46:03Z

1

I'm doing it like this:

value=`eval echo \\${IS_${i}}`

There's probably a better way but this should work.

edited Sep 11, 2017 at 11:46

answered Jun 22, 2017 at 7:30

papkass

1,2992 gold badges14 silver badges20 bronze badges

Collectives™ on Stack Overflow

Variable substitution in Shell script

2 Answers 2

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related