how to expand variables within a variable in Shell

Question

I got stuck at a problem where I need to expand a variable in another variable as follow :

var1=abc
var2=$var1/pqr
echo ${!var2}

Here I want output to be abc/pqr , but not getting it , please help. and var2 value is like this only , I can't put it in double quotes.

R Sahu · Accepted Answer · 2014-04-18 06:01:04Z

1

>> a="abcd"
>> b='$a/xyz'
>> eval c="$b"

>> echo $b
   $a/xyz
>> echo $c
   abcd/xyz

answered Apr 18, 2014 at 6:01

R Sahu

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2 Comments

eval is certainly not indicated here.

Even in such tiny examples it would be a good idea to teach people to use quotes correctly.

merlin2011 · Accepted Answer · 2014-04-18 05:59:38Z

1

Just remove the !.

var1=abc
var2=$var1/pqr
echo ${var2}

Note the curly braces are not necessary above, but they do not hurt either.

answered Apr 18, 2014 at 5:59

merlin2011

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jaypal singh · Accepted Answer · 2014-04-18 06:04:36Z

1

You don't need de-referencing in this case:

$ var1=abc
$ var2=$var1/pqr
$ echo $var2
abc/pqr

You de-reference when you assign variable as the value without the $ sigil. For example:

$ var1=abc/pqr
$ var2=var1
$ echo ${!var2}
abc/pqr

answered Apr 18, 2014 at 5:59

jaypal singh

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here I am getting output as below : echo $var2 : o/p $var1/pqr

I am not using any kind of quotes.