BASH using double variable $ - bad substitution [duplicate]

Question

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Bash - variable variables [duplicate] (4 answers)

Dynamic variable names in Bash (19 answers)

Closed 4 years ago.

From my code below, how to make the value of 'zz' become 500 after replacing 'critical_' with x on variable 'yy'

xab123=500

yy="critical_ab123"
zz=${"${yy//critical_/x}"}

echo $zz

instead the result, there is an error:

line 8: ${"${yy//critical_/x}"}: bad substitution

thanks adi

Answer 1

May be like this:

xab123=500
yy="critical_ab123"
zz="${yy//critical_/x}"
echo ${!zz}
500

Answer 2

An interesting usage is when you call a bash function, you can use indirection on the parameters passed in. Then, you can nest calls to your indirection function in a nested manner using command substitution.

deref() { echo "${!1}"; }

aa="bb"
bb="cc"
cc="hello"

echo "$(deref aa)" # bb
echo "$(deref "$(deref aa)")" # cc
echo "$(deref "$(deref "$(deref aa)")")" # hello

Here's deref used to solve the OP's problem:

deref() { echo "${!1}"; }

xab123="500"
yy="critical_ab123"
zz="$(deref "${yy//critical_/x}")"

echo "$zz" # Outputs: 500

Applied edits based on @charles-duffy comments:

1. Disclaimer: reader beware, there are performance impacts to the command substitution used in this approach (FIFO creation, fork() of a subshell, read() and wait().
2. Quotes were added to protect against lossy expansion, i.e. echo "$zz" is better than echo $zz
3. Use POSIX-compliant function declaration syntax, i.e. replaced function deref { echo "${!1}" ; } with deref() { echo "${!1}" ; }
4. Corrected quoting issue for each quoting context