How to use double $ in echo [duplicate]

Question

[sudhir.tataraju.WBLL0717E4418] ➤ cat test.sh

#!bin/bash

i=2

value2=abc

echo "`$value$i`"

Output:

[sudhir.tataraju.WBLL0717E4418] ➤ sh test.sh
test.sh: line 8: 2: command not found

Debug output:

[sudhir.tataraju.WBLL0717E4418] ➤ sh -x test.sh
+ i=2
+ value2=abc
++ 2
test.sh: line 8: 2: command not found
+ echo ''

How to get output abc without error?

Answer 1

What you are trying to do can be achieved much better with arrays.

values=( value1 value2 value3 "This is value 4" )
echo "${values[2]}"
values[2]="other_value"
echo "${values[2]}"

Variables can also be used as the index:

i=2
echo "${values[$i]}"

---

Arrays make looping over the values very easy:

#looping over values
for value in "${values[@]}"; do
    echo "$value"
done
#looping over indexes
for index in "${!values[@]}"; do
    echo "${values[$index]}"
done

If you don't consider sparse or associative arrays, you can also loop like this:

for (( index=0; index<${#values[@]}; index++ )); do
    echo "${values[index]}"
done

---

Arrays are the right structure for your task. They offer great functionality. For more information on arrays, see: wiki.bash-hackers.org/syntax/arrays and Bash Reference Manual.

Answer 2

What you're asking for can be achieved in two ways (without antipatterns or eval).

1. Use indirect expansion:

   The basic form of parameter expansion is ${parameter}. The value of parameter is substituted. [...] If the first character of parameter is an exclamation point (!), [...] it introduces a level of variable indirection. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion.

   So what you want is:

   #!/bin/bash
   
   i=2
   
   value2=abc
   
   # Have a variable that expands to the name of the variable you want
   myvar=value$i
   
   # Use indirect expansion
   echo "${!myvar}"
   

2. Realize that your design is very bad, and use some proper structure instead: Bash handles arrays. See PesaThe's answer. That's definitely the recommended method in your case.

---

Don't use eval. Though it might look great, you need a good deal of practice of shell scripting before you understand all the problems with eval. And when you have enough experience with Bash, you'll realize that you very very rarely need eval.