Find contents of parentheses that start with UPPERCASE letter?

Question

I want to grab the contents of parentheses, excluding the parentheses. Then add a colon to the end plus font items shown below.

Before: (Woman 1), (Ki-Woo), (Drunk)

After: <font color="#FF4500"><b>Woman 1:</b></font>

Here's what I have so far:

Find: (\([A-Z]*(?:(\h*|-)[A-Z0-9][a-z]*)*\))

Replace: \<font color\=\"\#FFA500\"\>\<b\>($1)\:\<\/b\>\<\/font\>

Currently mine still includes the brackets in the Find.

Answer 1

Using the pattern that you tried, you can switch the capture group to the inner part of the parenthesis, and make the second capture group a non capturing one.

Then use group 1 in the replacement.

\(([A-Z]*(?:(?:\h*|-)[A-Z0-9][a-z]*)*)\)

Regex demo

To not match empty parenthesis, you might write the pattern as:

\(([A-Z][a-z]*(?:(?:\h*|-)[A-Z0-9][a-z]*)*)\)

• \( Match (
• ( Capture group 1
  • [A-Z][a-z]* Match a single char A-Z and optional chars a-z
  • (?: Non capture group to repeat as a whole part
    • (?:\h*|-) Match either 0+ horizontal whitspace chars or a single -
    • [A-Z0-9][a-z]* Match a single char A-Z and optional chars a-z
  • )* Close and repeat the non capture group 0+ times
• ) Close group 1
• \) Match )

Regex demo

Note that this part of the pattern (?:\h*|-) can match either optional horizontal whitespace chars, or a single hyphen, and matches Ki Woo but could also match KiWoo

In the replacement use group 1 using $1

<font color="#FF4500"><b>$1</b></font>