concatenate variables in a bash script

Question

Hi I would like to set a variable by concatenating two other variables.

Example

A=1
B=2
12=C

echo $A$B

desired result being C

however the answer I get is always 12

Is it possible?

UPDATED Example

A=X
B=Y
D=$A$B

xy=test

echo $D

desired result being "test"

12 is not a variable, and there's no reason why it should map automatically? — Pieter21
– Pieter21, Commented Sep 22, 2014 at 16:30
can't 12 be a variable? I'm only using it as an example. What if we changed 12 to XY — user3246938
– user3246938, Commented Sep 22, 2014 at 16:40
To be clearer (and maybe get a better answer), I suggest you change the name of this question to "Look up bash variable name by value", which is the more difficult part of what you're trying to do. — Jeff Bowman
– Jeff Bowman, Commented Sep 22, 2014 at 16:41
I don't think you can look up a variable by value that easy. Suppose both 'C' and 'XY' have value 12, and you'd want to lookup 12. What would be the desired outcome? Any one? A collection? The most recent one? — Pieter21
– Pieter21, Commented Sep 22, 2014 at 16:42

Etan Reisner · Accepted Answer · 2014-09-22 16:38:18Z

2

It looks like you want indirect variable references.

BASH allows you to expand a parameter indirectly -- that is, one variable may contain the name of another variable:
# Bash
realvariable=contents
ref=realvariable
echo "${!ref}"   # prints the contents of the real variable

But as Pieter21 indicates in his comment 12 is not a valid variable name.

answered Sep 22, 2014 at 16:38

Etan Reisner

81.6k8 gold badges119 silver badges154 bronze badges

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

2 Comments

user3246938 Over a year ago

Thanks for the reply but this is not what I am trying to do. I'm trying to call a varible by using varibles

Etan Reisner Over a year ago

Yes, that's what this does. It lets you construct a variable whose contents are the name of another variable to get the value from. Try realvariable=contents; a=real; b=variable; ref=$a$b; echo "${!ref}".

legends2k · Accepted Answer · 2014-09-22 16:42:21Z

2

Since 12 is not a valid variable name, here's an example with string variables:

> a='hello'
> b='world'
> declare my_$a_$b='my string'
> echo $my_hello_world
my string

edited Sep 22, 2014 at 16:42

answered Sep 22, 2014 at 16:30

legends2k

33.2k28 gold badges131 silver badges238 bronze badges

2 Comments

user3246938 Over a year ago

No that's no what I am trying to do. I'm just trying to call $12 by creating the variable name from two other variables

legends2k Over a year ago

I think you need a dynamic variable, check the answer now.

bryn · Accepted Answer · 2014-09-22 17:32:17Z

What you are trying to do is (almost) called indirection: http://wiki.bash-hackers.org/syntax/pe#indirection

...I did some quick tests, but it does not seem logical to do this without a third variable - you cannot do concatenated indirection directly as the variables/parts being concatenated do not evaluate to the result on their own - you would have to do another evaluation. I think concatenating them first might be the easiest. That said, there is a chance you could rethink what you're doing. Oh, and you cannot use numbers (alone or as the starting character) for variable names.

Here we go:

cake="cheese"

var1="ca"
var2="ke"

# this does not work as the indirection sees "ca" and "ke", not "cake". No output.
echo ${!var1}${!var2}

# there might be some other ways of tricking it to do this, but they don't look to sensible as indirection probably needs to work on a real variable.
# ...this works, though:
var3=${var1}${var2}
echo ${!var3}

Collectives™ on Stack Overflow

concatenate variables in a bash script

3 Answers 3

2 Comments

2 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related