How to replace text inside another string which matches a regex pattern - bash

Question

${[a-zA-Z0-9._:]*}
I have used the above pattern for grepping words similar to ${SomeText}. How can I replace 'SomeText' with some other string i.e ${SomeText} --> SomeString in my bash script?
Example:
file.txt
text text text ${SomeText1} text text ${SomeText2} text text text text text text
text text text text ${SomeText3} text text text text text text ${SomeText4} text
...
my script file:
SomeText1="foo"
SomeText2="bar"
..
I want to replace ${SomeText1} to $SomeText1 which will be replaced by foo.
Similarly
${SomeText2} > $SomeText2 > bar
text text text foo text text bar text text text text text text
text text text text baz text text text text text text qux text
...

Answer 1

You can try this script:
while IFS='=' read k v; do
v=$(tr -d '"' <<< $v)
sed -i.bak "s/\${$k}/$v/g" file.txt
done < script

Answer 2

You can use sed
echo '${SomeText}, SomeText' | sed "s/\${SomeText}/\${SomeOtherText}/g"
will output
${SomeOtherText}, SomeText
if you need to use variables in sed, do it like this
a="SomeText"
b="SomeOtherText"
echo '${SomeText}, SomeText' | sed "s/\${$a}/\${$b}/g"
The output will be the same as above
EDIT:
Better use arrays to store that strings
SomeText[1]="foo"
SomeText[2]="bar"
i=0
while [[ $i < $MAX_i ]]
do
((i++))
sed -i "s/\${SomeText[$i]}/${SomeText[$i]}/g"
done
If you really need your approach, you will need double evaluation
SomeText[1]="foo"
SomeText[2]="bar"
i=0
while [[ $i < $MAX_i ]]
do
((i++))
toRepl="SomeText"$i
scriptToCall=$(eval echo "s/\\\${$toRepl}/$`eval echo ${toRepl}`/g")
sed -i $scriptToCall file
done
So you iterate thru all indices (SomeText1, SomeText2, ...), save the current into var toRepl, evaluate the script you want to call, for i=1 it will be scriptToCall="s/\${SomeText[1]}/foo/g" cause you double evaluate toRepl -> ${SomeText1} -> foo and you pass that script to sed :)
Is it clear now?