How to use a variable's value as another variable's name in bash [duplicate]

I want to declare a variable, the name of which comes from the value of another variable, and I wrote the following piece of code:

a="bbb"
$a="ccc"

but it didn't work. What's the right way to get this job done?

eval is used for this, but if you do it naively, there are going to be nasty escaping issues. This sort of thing is generally safe:

name_of_variable=abc

eval $name_of_variable="simpleword"   # abc set to simpleword

This breaks:

eval $name_of_variable="word splitting occurs"

The fix:

eval $name_of_variable="\"word splitting occurs\""  # not anymore

The ultimate fix: put the text you want to assign into a variable. Let's call it safevariable. Then you can do this:

eval $name_of_variable=\$safevariable  # note escaped dollar sign

Escaping the dollar sign solves all escape issues. The dollar sign survives verbatim into the eval function, which will effectively perform this:

eval 'abc=$safevariable' # dollar sign now comes to life inside eval!

And of course this assignment is immune to everything. safevariable can contain *, spaces, $, etc. (The caveat being that we're assuming name_of_variable contains nothing but a valid variable name, and one we are free to use: not something special.)

You can use declare and !, like this:

John="nice guy"
programmer=John
echo ${!programmer} # echos nice guy

Second example:

programmer=Ines
declare $programmer="nice gal"
echo $Ines # echos nice gal

This might work for you:

foo=bar
declare $foo=baz
echo $bar
baz

or this:

foo=bar
read $foo <<<"baz"
echo $bar
baz

You could make use of eval for this.
Example:

$ a="bbb"
$ eval $a="ccc"
$ echo $bbb
ccc

Hope this helps!

If you want to get the value of the variable instead of setting it you can do this

var_name1="var_name2"
var_name2=value_you_want
eval temp_var=\$$var_name1
echo "$temp_var"

You can read about it here indirect references.