Remove the Parenthesis and keep with contents within

Hello All,
This is my first post here in the forum and relatively new to UiPath although I’ve been a vivid user of the forum :wink:

This is my doubt,

  1. I have a string variable (it might be a combination of alphanumeric with special characters - for ex : “#H3ll0&@Ll26”)
    And this above mentioned string will be getting in the below fashion
    Type 1 - (#H3ll0&@Ll26):
    Type 2 - ((#H3ll0&@Ll26)):
    Type 3 - (#H3ll0(&)@Ll26):

So what I am looking for is, when I get a string like Type 1 - I need to manipulate it in such a way I should remove the open and closing parenthesis and make the string look like #H3ll0&@Ll26

For Type 2 - it should look like #H3ll0&@Ll26

For Type 3 - #H3ll0(&)@Ll26 (here the open & closing parenthesis inside the string should remain as such.

I’ve used the following techniques so far
a. regex - (?<=()[A-Za-z0-9!#$%^&_-@ ](?<!)) = when I use this, if there is a second open and closing parenthesis, it will create one null match which I don’t need.
b. remove & use substring

but these options not really helps my needs !

Can somebody please help me out with getting a proper solution ?

Thanks in advance :slight_smile:


Is colon at the end of the string content of string?

If so, the following will work.


If not, can you try the following?




Thank You so very much for the solution !

I would need both your resolutions because sometimes there might be a “special character” after the parenthesis … I have utilized both your solution into my flow.
It worked as expected :smiley:

It would be extremely helpful If I could get a solution for the below scenarios
For the below one’s I have to keep the parenthesis

  1. Test(12)Try) - for this one, all the open and close parenthesis needs to kept in place.
  2. Test(12Provide
  3. (Test)12 Provide

for all these 3 types, need to keep them and not replace or remove the parenthesis


Can you try the following expression?



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