Resolve bash variable containted in another variable [duplicate]

This question already has answers here: Bash expand variable in a variable (5 answers) 开发者_JS百科 Closed 2 years ago.

I have code like that:

TEXT_TO_FILTER='I would like to replace this $var to proper value
                in multiline text'
var=variable

All I want to get is:

TEXT_AFTER_FILTERED="I'd like to replace this variable to proper value"

So I did:

TEXT_AFTER_FILTERED=`eval echo $TEXT_TO_FILTER`
TEXT_AFTER_FILTERED=`eval echo $(eval echo $TEXT_TO_FILTER)`

Or even more weirder things, but without any effects. I remember that someday I had similar problem and I did something like that:

cat << EOF > tmp.sh
echo $TEXT_TO_FILTER
EOF
chmod +x tmp.sh
TEXT_AFTER_FILTERED=`. tmp.sh`

But this solution seems to be to much complex. Have any of You heard about easier solution?

---

For security reasons it's best to avoid eval. Something like this would be preferable:

TEXT_TO_FILTER='I would like to replace this %s to proper value'
var=variable
printf -v TEXT_AFTER_FILTERED "$TEXT_TO_FILTER" "$var"
# or TEXT_AFTER_FILTERED=$(printf "$TEXT_TO_FILTER" "$var")
echo "$TEXT_AFTER_FILTERED"

---

TEXT_AFTER_FILTERED="${TEXT_TO_FILTER//\$var/$var}"

or, using perl:

export var
TEXT_AFTER_FILTERED="$(echo "$TEXT_TO_FILTER" | perl -p -i -e 's/\$(\S+)/$ENV{$1} || $&/e')"

This is still more secure than eval.