[SOLVED] Create a variable based on arguemnt passed to a function

Issue

Please take some time to help me with this part(shell scripting)
I have a code that is

helper()
{
  var1="$1"
  #multiple lines of code
  .
  . 
  #so the variable must be created based on what value is passed as an argument.
  $var1_var_name="some string"
  #so I should be able to make a varible abc_var_name in this case

}
helper "abc"

So I want to create a variable inside helper function such that that variable name would be

if helper "abc" is called then the variable should be abc_var_name="some string"
or if helper "def" is called then the variable should be def_var_name="some string"

Solution

You probably want namerefs (with a recent enough version of bash):

$ helper() {
    declare -n tmp="$1"_var_name
    tmp="some string"
  }
$ helper "abc"
$ echo "$abc_var_name"
some string

If your version of bash is too old you can also use read to set a named variable:

$ helper() {
    IFS= read -r "$1"_var_name <<< "some string"
  }
$ helper "abc"
$ echo "$abc_var_name"
some string

Or printf:

$ helper() {
    printf -v "$1"_var_name '%s' "some string"
  }
$ helper "abc"
$ echo "$abc_var_name"
some string

For completeness, as suggested in comments, if you accept to slightly change your approach, associative arrays are another option: all your named variables become elements of associative array var_name, the keys of the array are the names you pass to your function, and the values are whatever you want. One important benefit is that you can easily loop over all your variables:

$ declare -A var_name=()
$ helper() {
    var_name["$1"]="some string"
  }
$ helper "abc"
$ helper "def"
$ for key in "${!var_name[@]}"; do
    printf '%s: %s\n' "$key" "${var_name[$key]}"
  done
def: some string
abc: some string

Answered By - Renaud Pacalet

Answer Checked By - Marilyn (WPSolving Volunteer)