Replace string inside curvy brackets

Question

I would need to replace the strings contained within the curved brackets with the same strings but with an initial prefix and curly brackets \fill{(test_string)}. Is this possible?
Example:

Lorem ipsum dolor sit amet, consectetuer adipiscing elit, sed diam
nonummy nibh euismod tincidunt ut laoreet dolore.
(first_string)
Lorem ipsum dolor sit amet, consectetuer adipiscing elit, sed diam nonummy
nibh euismod tincidunt ut laoreet dolore.
(second_string)

Transform in:

Lorem ipsum dolor sit amet, consectetuer adipiscing elit, sed diam
nonummy nibh euismod tincidunt ut laoreet dolore.
\fill{(first_string)}
Lorem ipsum dolor sit amet, consectetuer
adipiscing elit, sed diam nonummy nibh euismod tincidunt ut laoreet dolore.
\fill{(second_string)}

Thank you for your reply! I am interested in understanding and learning. Any useful links? — Andy Toff, Nov 11 '22 at 13:06
read the MDN pages on Regex, they deal with the JavaScript dialect used by VSC, there are tons of webpages about regex, read SO posts with tag `regex` — rioV8, Nov 11 '22 at 15:23

score 1 · Accepted Answer · answered Nov 11 '22 at 15:31

As suggested, you could use a regex search and replace. A simple example of this can be found here. In your case, this should work:

The regular expression $([^$]+)\) does the following (as taken from this site - you'll need to paste the regex into the site to see the explanation):

$ matches the character literally (case sensitive)

1st Capturing Group ([^$]+)

Match a single character not present in the list below [^\)]

+ matches the previous token between one and unlimited times, as many times as possible, giving back as needed (greedy)

\) matches the character ) literally (case sensitive)

\) matches the character ) literally (case sensitive)

In Visual Studio Code, if you enable regex search by clicking the .* icon in the search bar, you can put this regular expression in. Then, in the replace section, you can put \fill{($1)} where the $1 is the 1st Capturing Group mentioned previously (the first_string, second_string, etc. part found by the regular expression).

There are a lot of Regex posts here on Stackoverflow you may want to read. One notable one is Greedy versus Lazy.

You can use the simpler `($.+?$)` and replace with `\fill{$1}`. The greedy `?` simplifies it. And it is worth mentioning that nested `()` with a string is a big problem for a regex to solve. — Mark, Nov 11 '22 at 16:36

Replace string inside curvy brackets

1 Answers1