Bash – How to assign space-containing values to variables in bash using eval

bashquotingshell

I want to dynamically assign values to variables using eval.
The following dummy example works:

var_name="fruit"
var_value="orange"
eval $(echo $var_name=$var_value)
echo $fruit
orange

However, when the variable value contains spaces, eval returns an error, even if $var_value is put between double quotes:

var_name="fruit"
var_value="blue orange"
eval $(echo $var_name="$var_value")
bash: orange : command not found

Any way to circumvent this ?

Best Answer

Don't use eval, use declare

$ declare "$var_name=$var_value"
$ echo "fruit: >$fruit<"
fruit: >blue orange<

Related Solutions

Bash – Read non-bash variables from a file into bash script

While you can transform this file to be a shell snippet, it's tricky. You need to make sure that all shell special characters are properly quoted.

The easiest way to do that is to put single quotes around the value and replace single quotes inside the value by '\''. You can then put the result into a temporary file and source that file.

script=$(mktemp)
sed <"config-file" >"$script" \
  -e '/^[A-Z_a-z][A-Z_a-z]*=/ !d' \
  -e s/\'/\'\\\\\'\'/g \
  -e s/=/=\'/ -e s/\$/\'/

I recommend doing the parsing directly in the shell instead. The complexity of the code is about the same, but there are two major benefits: you avoid the need for a temporary file, and the risk that you accidentally got the quoting wrong and end up executing a part of a line as a shell snippet (something like dangerous='$(run me)'). You also get a better chance at validating potential errors.

while IFS= read -r line; do
  line=${line%%#*}  # strip comment (if any)
  case $line in
    *=*)
      var=${line%%=*}
      case $var in
        *[!A-Z_a-z]*)
          echo "Warning: invalid variable name $var ignored" >&2
          continue;;
      esac
      if eval '[ -n "${'$var'+1}" ]'; then
        echo "Warning: variable $var already set, redefinition ignored" >&2
        continue
      fi
      line=${line#*=}
      eval $var='"$line"'
  esac
done <"config-file"

Bash – Single Quote Within Double Quotes and the Bash Reference Manual

The ' single quote character in your echo example gets it literal value (and loses its meaning) as it enclosed in double quotes ("). The enclosing characters are the double quotes.

What you can do is print the single quotes separately:

echo "'"'$a'"'"

or escape the $:

echo "'\$a'"

Best Answer

Related Solutions

Bash – Read non-bash variables from a file into bash script

Bash – Single Quote Within Double Quotes and the Bash Reference Manual

Related Question