Replacement of strings within 2 strings in regex - regex

Question

I have a string:
dkj a * & &*(&(*(
//#HELLO
^%#&UJNWDUK()C*(v 8*J DK*9
//#HE#$^&&(akls#$98akdjl ak##sjdkja
//
%^&*(//#HELLO//#BYE<><>
//#BYE
^%#&UJNWDUK()C*(v 8*J DK*90K )
//#HELLO
&*^J$XUK 8j8 j jk kk8(&*(
//#BYE
and I need to have 2 groups such as each group must start with //HELLO then there should be a next line and any type of text can follow (.*) but it will end with a //BYE preceded by a line:
1)
//#HELLO
^%#&UJNWDUK()C*(v 8*J DK*9
//#HE#$^&&(akls#$98akdjl ak##sjdkja
//
%^&*(//#HELLO//#BYE<><>
//#BYE
2)
//#HELLO
&*^J$XUK 8j8 j jk kk8(&*(
//#BYE
and replaces the original string to this: (basically adding // to each line of each group)
dkj a * & &*(&(*(
////#HELLO
//^%#&UJNWDUK()C*(v 8*J DK*9
////#HE#$^&&(akls#$98akdjl ak##sjdkja
////
//%^&*(//#HELLO//#BYE<><>
////#BYE
^%#&UJNWDUK()C*(v 8*J DK*90K )
////#HELLO
//&*^J$XUK 8j8 j jk kk8(&*(
////#BYE
Here is my current progress:
I have
\/\/#HELLO\n.*?\/\/#BYE[\n$]
However im not sure how to go about the replacement, I'm thinking separating each line per group using \G after the //#HELLO and ending with //#BYE

Answer 1

It's a bit complex, but this will do it:
Search: (?m)(//#HELLO[\r\n]+|\G(?://#BYE|(?=(?:[^#]|#(?!HELLO[\r\n]+))*#BYE)[^\r\n]*[\r\n]*))
Replace: //$1
In Groovy:
String resultString = subjectString.replaceAll(/(?m)(\/\/#HELLO[\r\n]+|\G(?:\/\/#BYE|(?=(?:[^#]|#(?!HELLO[\r\n]+))*#BYE)[^\r\n]*[\r\n]*))/, '//$1');

Answer 2

For grouping into separate lines use the following regex:
//#HELLO\r(.*[\n\r]+)*//#BYE\r?
\r - Newline character
[\n\r] - Enter characters
*? - Non-greedy match
?- Match 1 or 0 times
You can take out the ? at the end if it always ends with a newline.
You can then use the group (The value inside the brackets) to search and replace.