shell scripting - assigning one variable to another - arrays

Question

I have a doubt. When i declare a value and assign to some variable, I don't know how to reassign the same value to another variable. See the code snippet below.

Here is my actual script.

#!/bin/sh

a=AA
b=BB
c=CC
d=DD
e=EE
f=FF

alpha_array=(a b c d e f)
process_array=(proc1 proc2 proc3 proc4)
array_1=("")
array_2=("")

display_array() {
echo "array1 = ${array_1[@]}"
echo "array2 = ${array_2[@]}"
}

checkarg() {
if [[ " ${alpha_array[*]} " == *" $token "* ]]; then
    echo "alphabet contains $token "
    array_1=("${array_1[@]}" "$token")
    $token=${$token}
    echo "TOKEN = $token"
elif [[ " ${process_array[*]} " == *" $token "* ]]; then
    echo "process contains $token "
    array_2=("${array_2[@]}" "$token")
else
echo "no matches found"
display_array
exit 1
fi
}

for token in $@
do
   echo $token
   checkarg
done

display_array

Here the below two lines

$token=${$token}
echo "TOKEN = $token"

should display my output as

TOKEN = AA
TOKEN = BB

when i run my script with the following arguments.

./build.sh a b proc1

Kindly help me out on those 2 lines.

What do you want the line $token=${$token} to do? That is, if it operated correctly, what would it accomplish? — Charles Duffy
– Charles Duffy, Commented Dec 19, 2014 at 0:59
By the way, the modern replacement for array_1=("${array_1[@]}" "$token") is array_1+=( "$token" ) -- both shorter to write and faster execution. — Charles Duffy
– Charles Duffy, Commented Dec 19, 2014 at 1:01
I need to assign the incoming argument 'a' to token and get the value of $a. — Murthy
– Murthy, Commented Dec 19, 2014 at 1:03

Charles Duffy · Accepted Answer · 2014-12-19 01:04:14Z

2

It sounds like what you want is variable indirection. There's a lot of code in your question which has nothing to do with that, but let me try to distill it down to what I understand as the important parts:

a=AA
b=BB
alpha_array=(a b)
for token in "${alpha_array[@]}"; do
  value=${!token}
  echo "value of variable $token is $value"
done

When run, this will output:

value of variable a is AA
value of variable b is BB

For more details, see BashFAQ #6.

By the way, this can often be replaced with use of associative arrays, in which case you might write:

declare -A tokens=( [a]=AA [b]=BB )
for token in "${!tokens[@]}"; do
  value=${tokens[$token]}
  echo "value of token $token is $value"
done

This has the advantage that your key/value pairs are all stored inside the array -- there's no potential for collision with other variable names.

answered Dec 19, 2014 at 1:04

Charles Duffy

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