Bash: Find and replace using Parameter Expansion

Question

I want to replace the input,

find_string: @include circle-progress(38px, 30px, #4eb630)

and output,

Output_string: @include circle-progress(38px, 30px)

using ${find_string//pattern/replacement_string} where pattern is , , #[A-Fa-f0-9][A-Fa-f0-9][A-Fa-f0-9]?([A-Fa-f0-9]?([A-Fa-f0-9]?([A-Fa-f0-9])))' that I supply.

In the code below, simply the line matching pattern is printed i.e find_string, when I read lines of code from a file, whereas I want the output_string to be printed.

pattern="@include circle-progress\(([0-9]{1,3}px, ){2}#[A-Fa-f0-9]
{3,6}\)" /*regex the matches find_string*/

replace_glob=', #[A-Fa-f0-9][A-Fa-f0-9][A-Fa-f0-9]?([A-Fa-f0-9]?([A-
Fa-f0-9]?([A-Fa-f0-9])))' /*glob pattern in the string to be replaced*/

while IFS='' read -r line || [[ -n "$line" ]]; do
   if [[ $line =~ $pattern ]] 
   then
      echo "${line//$replace_glob/}" 
   fi
done < "$1"

Answer 1

The pattern in parameter expansion is not a regular expression but follows the same rules as glob pattern matching:

• * : matches any character sequence
• ? : matches any character
• [..] : any character in set
• [^..] or [!..] : any character not in set

with shell option : shopt -s extglob, some more features but less than regular expressions

• @(..|..) : match any once
• ?(..|..) : match any 0 or 1 times
• *(..|..) : match any 0 or more times
• !(..) : matches all except

However bash supports some basic regex, following should work:

string='@include circle-progress(38px, 30px, #4eb630)'
pattern='@include circle-progress\([ ]*[0-9]{1,3}px,[ ]*[0-9]{1,3}px(,[ ]*#[A-Fa-f0-9]{3,6}[ ]*)\)'
[[ $string =~ $pattern ]] && echo "${string//"${BASH_REMATCH[1]}"}"