How do I remove multiple bracket within nested bracket using sed? - regex

Question

I want to make a program editing file code with shell programming
there is command 'Remove ${} in arithmetic expansion$(())'
and I have a problem in implementation.
I'm going to make bash shell code below
cnt=$(( ${cnt} + ${cnt123} ))
to
cnt=$(( cnt + cnt123 ))
I want to remove command substitution bracket in arithmetic Expansion bracket
I tried to do with this regex expression:
sed -Ei 's/(\$\(\()([^\)]*?)\$\{([^\}]+?)\}(.*?)(\)\))/\1\2\3\4\5/g' $file
but, it just found a longest one.(even though there are another match before matched one)
if you want to see visualized regex expression, click this link visualized image
result showed like this:
cnt=$(( ${cnt} + cnt123 ))
How to do remove internal bracket in nested bracket?
(I should just use awk or sed, but if it could be impossible, it doesn't matter using other bash command)
works example:
s=$(( ${s} ** 2 ))
to
s=$(( s ** 2 ))
sum=$(( ${a} + ${b} ))
to
sum=$(( a + b ))
echo $(( (${var} * ${var2}) / ${var3} ))
to
echo $(( (var * var2) / var3 ))
echo ${d} $((${t1} + ${t2})) ${e}
to
echo ${d} $(( t1 + t2 )) ${e}
:my sample input file (it doesn't matter what it do for)
#! /bin/bash
cnt=0
cnt123=1
for filename in *
do
fname=$(basename $filename)
cname=$(echo $fname | tr A-Z a-z)
if [ "$fname" != "$cname" ]
then
if [ -e "$cname" ]
then
echo "$cname already exists"
exit 1
fi
echo "$fname is renamed $cname"
mv $fname $cname
cnt=$(( ${cnt}+ ${cnt123} ))
fi
done
echo "Total count: $cnt"
exit 0

Answer 1

As it is not an easy task for sed/awk regex and relevant functions,
please let me show a perl solution, although perl is not tagged in
your question.
perl -i -pe 's/(?<=\$\(\().+?(?=\)\))/ $_ = $&; s#\$\{(.+?)}#$1#g; $_ /ge' "$file"
Input file example:
s=$(( ${s} ** 2 ))
sum=$(( ${a} + ${b} ))
echo $(( (${var} * ${var2}) / ${var3} ))
echo ${d} $((${t1} + ${t2})) ${e}
Modified result:
s=$(( s ** 2 ))
sum=$(( a + b ))
echo $(( (var * var2) / var3 ))
echo ${d} $((t1 + t2)) ${e}
The perl substitution s/pattern/expression/e replaces the matched
pattern with the perl expression instead of literal (or fixed)
replacement. You can perform a dynamic replacement with this mechanism.
(?<=\$\(\().+?(?=\)) matches a substring which is preceded by $((
and followed )). Then the matched substring will be the content within
$(( .. )). The perl variable $& is assigned to it.
The expression $_ = $&; s#\$\{(.+?)}#$1#g; $_ is a perl code to remove
the paired ${ and } from the matched substring above. The g option
after the delimiter # works as that of sed.

Answer 2

Depending on the context you may want to limit the extent of the match to only alphanumeric chars
sed -Ei.bak 's/\$\{([[:alnum:]]+)\}/\1/g'
to avoid unintentionally matching something else.

Answer 3

Using sed
$ sed -Ei.bak 's/\$\{([^}]*)}/\1/g' input_file
cnt=$(( cnt + cnt123 ))

Answer 4

printf="cnt=\$(( \${cnt} + \${cnt123} ))" | tr -d '{' | tr -d '}'
I used printf here because it is slightly more trustworthy than echo. Make sure you escape $, otherwise you will get an error because of bash trying to interpret the variables.