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.
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.
\(([A-Z]*(?:(?:\h*|-)[A-Z0-9][a-z]*)*)\)
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 )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>
Short code as always is below, just checked by me, Working perfect...
Find: \(([\w \d-]*)\)
Replace All: \<font color\=\"\#FFA500\"\>\<b\>$1\:\<\/b\>\<\/font\>