bash while in while using variable_variable operation

Question

I'm trying to create a loop menu script and my main-while turn to false when secondary menu turn to false... what do you think that I need fix?

#!/bin/bash

function f_menu_main () {
  echo "1) option1";
  echo "2) option2";
  echo "3) option3";
  echo "4) option4";
  echo "99) exit";
}

function f_case_main () {
  case $selection0 in
    "1" ) function1; selection0="1"; ;;
    "2" ) function2; selection0="2"; ;;
    "3" ) function3; selection0="3"; ;;
    "4" ) f_run_app packages 1; ;;
    "99" | "q" | "exit" | "quit") selection0="exitl0"; ;;
    *) f_menu_main; ;;
  esac
}

function f_case_packages () {
  case $selection1 in
    "1" ) function1; selection1="1"; ;;
    "2" ) function2; selection1="2"; ;;
    "99" | "q" | "exitl" | "quit") selection1="exitl1"; ;;
    *) f_menu_packages; ;;
  esac
}

function f_menu_packages () {
  echo "1)  options";
  echo "2) options";
  echo "99) exit";
}

function f_run_app () {
  selection="selection"$2;
  exitlv="exitl"$2;
  while [ "${!selection}" != "$exitlv" ]; do
    echo "";
    f_menu_$1;
    echo "Last selection: \""${!selection}" "$2"\".";
    echo -n "Select a item from menu: "; read "selection"$2;
    f_case_$1;
  done
}

f_run_app main 0;

My supposition is when selection1 is exit1 while1 exit but while0 exit too and the variable selection0 is not exit0.

Answer 1

Try placing your variables in local context and see if that would help. The subcall probably modifies it and doesn't put it back to the original value after exiting.

function f_run_app () {
  local selection="selection"$2;
  local exitlv="exitl"$2;
  while [ "${!selection}" != "$exitlv" ]; do
    echo "";
    f_menu_$1;
    echo "Last selection: \""${!selection}" "$2"\".";
    echo -n "Select a item from menu: "; read "selection"$2;
    f_case_$1;
  done
}