Ruby Regex Group Replacement

Question

I am trying to perform regular expression matching and replacement on the same line in Ruby. I have some libraries that manipulate strings in Ruby and add special formatting characters to it. The formatting can be applied in any order. However, if I would like to change the string formatting, I want to keep some of the original formatting. I'm using regex for that. I have the regular expression matching correctly what I need:

mystring.gsub(/[(\e\[([1-9]|[1,2,4,5,6,7,8]{2}m))|(\e\[[3,9][0-8]m)]*Text/, 'New Text')

However, what I really want is the matching from the first grouping found in:

(\e\[([1-9]|[1,2,4,5,6,7,8]{2}m))

to be appended to New Text and replaced as opposed to just New Text. I'm trying to reference the match in the form of

mystring.gsub(/[(\e\[([1-9]|[1,2,4,5,6,7,8]{2}m))|(\e\[[3,9][0-8]m)]*Text/, '\1' + 'New Text')

but my understanding is that \1 only works when using \d or \k. Is there any way to reference that specific capturing group in my replacement string? Additionally, since I am using an asterik for the [], I know that this grouping could occur more than once. Therefore, I would like to have the last matching occurrence yielded.

My expected input/output with a sample is:

Input:  "\e[1mHello there\e[34m\e[40mText\e[0m\e[0m\e[22m"
Output: "\e[1mHello there\e[40mNew Text\e[0m\e[0m\e[22m"

Input:  "\e[1mHello there\e[44m\e[34m\e[40mText\e[0m\e[0m\e[22m"
Output: "\e[1mHello there\e[40mNew Text\e[0m\e[0m\e[22m"

So the last grouping is found and appended.

Answer 1

You can use the following regex with back-reference \\1 in the replacement:

reg = /(\\e\[(?:[0-9]{1,2}|[3,9][0-8])m)+Text/
mystring = "\\e[1mHello there\\e[34m\\e[40mText\\e[0m\\e[0m\\e[22m"
puts mystring.gsub(reg, '\\1New Text')

mystring = "\\e[1mHello there\\e[44m\\e[34m\\e[40mText\\e[0m\\e[0m\\e[22m"
puts mystring.gsub(reg, '\\1New Text')

Output of the IDEONE demo:

\e[1mHello there\e[40mNew Text\e[0m\e[0m\e[22m
\e[1mHello there\e[40mNew Text\e[0m\e[0m\e[22m

Mind that your input has backslash \ that needs escaping in a regular string literal. To match it inside the regex, we use double slash, as we are looking for a literal backslash.