How to substitute variables in read with it's value?-CodePudding

Here's the code:

#!/bin/bash

read
dir =("$REPLY")

echo "${dir[@]}"
for i in "${dir[@]}"; do
        echo "$i"
done

My question is: how to substitute $some_variable in read with it's value?

CodePudding user response：

You can use variable indirection with parameter expansion/substitution to replace variable names with their values.

read
while [[ $REPLY =~ \$([a-z_A-Z][a-z_A-Z0-9]*) ]] ; do
    var=${BASH_REMATCH[1]}
    REPLY=${REPLY//\$$var/${!var}}
done
dir =("$REPLY")

Note that if the variable contains a name of another variable, it will be expanded as well, ad infitum. You might want to prevent that somehow.

CodePudding user response：

You don't want eval, because that would try to change more than the variables.
You can use envsubst:

REPLY='$HOME/a/$(date)'
dir =("$REPLY")

echo "Translating ${dir[@]}"
for i in "${dir[@]}"; do
  echo "Original code"
  echo "  $i"
  echo "Wrong code using crazy eval"
  eval echo "  ${i}"
  echo "When you have envsubst"
  echo "  ${i}" | envsubst
done

Result:

Translating $HOME/a/$(date)
Original code
  $HOME/a/$(date)
Wrong code using crazy eval
/home/username/a/Mon Feb 7 22:51:58 CET 2022
When you have envsubst
  /home/username/a/$(date)