I want to create a variable from multiple variables by concatenating their values. After that I want to update value in those nested child variables and see the change in the parent one. Something like this:
#!/bin/bash
a=123
b=abc
c=$a$b
echo $c # outputs 123abc
a=456
echo $c # outputs 123abc although I want 456abc
I'd like this to output 123abc and 456abc but instead I get 123abc and 123abc. Is it possible to achieve the behaviour I want in bash?
In BASH (4+) you can create reference of variables using declare -n so if you have:
a=123
And create a variable reference as:
declare -n c=a
then echo $c will print 123.
If you now change to a=789 then if you execute echo $c again you will get updated value of 789.
However this reference can be created for a single variable (or array) only, nor for combination of multiple variables.
As a simple work around consider using shell function for your usecase as this:
c() { echo "$a$b"; }
Then $(c) will always be dynamic and would always get you update value by concatenating values a and b.
I think this is what you are looking for. Be careful though as eval use can be dangerous if you don't trust the input.
#!/bin/bash
a=123
b=abc
c="\${a}\${b}"
eval "echo $c"
a=456
eval "echo $c"
Hope it helps!
eval discouraged?
c=$a$b, you're not assigning references ofaandbtoc, but you're assigning their values.c() { echo "$a$b"; }and use it as$(c)in your script and you will get updated values all the time.