Bash variable substitution and strings - bash

Question

Let's say I have two variables:
a="AAA"
b="BBB"
I read a string from a file. This string is the following:
str='$a $b'
How to create a new string from the first one that substitutes the variables?
newstr="AAA BBB"

Answer 1

bash variable indirection whithout eval:
Well, as eval is evil, we may try to make this whithout them, by using indirection in variable names.
a="AAA"
b="BBB"
str='$a $b'
newstr=()
for cnt in $str ;do
[ "${cnt:0:1}" == '$' ] && cnt=${cnt:1} && cnt=${!cnt}
newstr+=($cnt)
done
newstr="${newstr[*]}"
echo $newstr
AAA BBB
Another try:
var1="Hello"
var2="2015"
str='$var1 world! Happy new year $var2'
newstr=()
for cnt in $str ;do
[ "${cnt:0:1}" == '$' ] && cnt=${cnt:1} && cnt=${!cnt}
newstr+=($cnt)
done
newstr="${newstr[*]}"
echo $newstr
Hello world! Happy new year 2015
Addendum As correctly pointed by #EtanReisner's comment, if your string do contain some * or other glob expendable stings, you may have to use set -f to prevent bad things:
cd /bin
var1="Hello"
var2="star"
var3="*"
str='$var1 this string contain a $var2 as $var3 *'
newstr=()
for cnt in $str ;do
[ "${cnt:0:1}" == '$' ] && cnt=${cnt:1} && cnt=${!cnt};
newstr+=("$cnt");
done;
newstr="${newstr[*]}"
echo "$newstr"
Hello this string contain a star as * bash bunzip2 busybox....zmore znew
echo ${#newstr}
1239
Note: I've added " at newstr+=("$cnt"); to prevent glob expansion, but set -f seem required...
newstr=()
set -f
for cnt in $str ;do
[ "${cnt:0:1}" == '$' ] && cnt=${cnt:1} && cnt=${!cnt}
newstr+=("$cnt")
done
set +f
newstr="${newstr[*]}"
echo "$newstr"
Hello this string contain a star as * *
Nota 2: This is far away from a perfect solution. For sample if string do contain ponctuation, this won't work again... Example:
str='$var1, this string contain a $var2 as $var3: *'
with same variables as previous run will render:
' this string contain a star as *' because ${!var1,} and ${!var3:} don't exist.
... and if $str do contain special chars:
As #godblessfq asked:
If str contains a line break, how do I do the substitution and preserve the newline in the output?
So this is not robust as every indirected variable must be first, last or space separated from all special chars!
str=$'$var1 world!\n... 2nd line...'
var1=Hello
newstr=()
set -f
IFS=' ' read -d$'\377' -ra array <<<"$str"
for cnt in "${array[#]}";do
[ "${cnt:0:1}" == '$' ] && cnt=${cnt:1} && cnt=${!cnt}
newstr+=("$cnt")
done
set +f
newstr="${newstr[*]}"
echo "$newstr"
Hello world!
... 2nd line...
As <<< inline string add a trailing newline, last echo command could be written:
echo "${newstr%$'\n'}"

Answer 2

The easiest solution is to use eval:
eval echo "$str"
To assign it to a variable, use command substitution:
replaced=$(eval echo "$str")

Answer 3

Disclaimer: I only discovered perl an hour ago. But this seems to work robustly, whatever special characters you throw at it:
newstr=$(a2="$a" b2="$b" perl -pe 's/\$a\b/$ENV{a2}/g; s/\$b\b/$ENV{b2}/g' <(echo -e "$str"))
Test:
a='A*A\nA'
b='B*B\nB'
str='$a $aa * \n $b $bb'
newstr=$(a2="$a" b2="$b" perl -pe 's/\$a\b/$ENV{a2}/g; s/\$b\b/$ENV{b2}/g' <(echo -e "$str"))
echo -e "$newstr"
Output:
A*A
A $aa *
B*B
B $bb

Answer 4

I'd use awk solution with awk-variables. This will allow passing a text containing special chars and subsitute any placeholder with it.
a workaround to recognize $ would be using [\x24]:
awk -v a="$a" -v b="$b" '{gsub("[\x24]a",a);gsub("[\x24]b",b); print}' <<< $str
here
-v defines variable a="$a"
[x24] is ASCII for $, so [x24]a equal to $a
gsub(x,y) - replaces x with y