How to make sure if every opening bracket has a closing one (with RegEx)

Hi

I need to make sure that every opening bracket in a long string has also been closed at the end or vice versa.
i.e
string="abc(xyz)...gsdfs(asdfasdfadfaf
this should put another brakcer at the end in order to make its pair.
There are code snippets with a loop but I want to achieve it with RegEx.is it possible
Reason I don't want to use loop ,is because I will have to loop over cells in a range and if every cell has a sub loop then the code will go slow.

Hi nilem
Good move
but it doesn't tells either "(" is missing or ")" is missing (and how many missing) because I need to complete the pairs as well.

This is a little clunky, but it will tell you if a string has the parenthesis positioned correctly.

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No routine can tell where a missing parenthesis goes. From your example:
abc(xyz)...gsdfs(asdfasdfadfaf is missing a right paren.
but is abc(xyz)...gsdfs(asdfasdfadfaf) the correct entry
or is abc(xyz)...gsdfs(asdfasdf)adfaf

Another function you might use is

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If ParenDifference is 0, there are matching left and right parenthesis. If ParenDifference is positive, then that is the number of excess left parenthesis. Negative is the number excess right parenthesis.

Note that this is only the count of parenthesis and their position is not a factor, ParenDifference( "abc(abc))xy(") = 0

Another way for the second:

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Another way for the first:

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If you want to ignore embedded substrings (e.g., ")"(ab)c is fine), and also verify that there are an even number of quotes (e.g., ")"(ab)c" is not),

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Originally Posted by mikerickson

No routine can tell where a missing parenthesis goes. From your example:
abc(xyz)...gsdfs(asdfasdfadfaf is missing a right paren.
but is abc(xyz)...gsdfs(asdfasdfadfaf) the correct entry
or is abc(xyz)...gsdfs(asdfasdf)adfaf

Hi Sir Mike
Thanks for reviewing the thread
The pattern is not important currently.What I need is to determine if there is ")" missing I will add it at the very end of the string and if "(" is less ,I will put it at the beginning of the string.
I want to avoid loop here because there will be another loop on the cells in my range separately . I am afraid if I use loop within each cell and then for each cell my code will loose robustness. because there will cells in thousands

Originally Posted by mikerickson

Another function you might use is

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If ParenDifference is 0, there are matching left and right parenthesis. If ParenDifference is positive, then that is the number of excess left parenthesis. Negative is the number excess right parenthesis.

Note that this is only the count of parenthesis and their position is not a factor, ParenDifference( "abc(abc))xy(") = 0

But ParenDifference("abc(xyz)...gsdfs(asdfasdfadfaf") = 1
ParenDifference("abc(xyz)...gsdfs(asdfasdfadfaf)))")= -1

Means

OK....I further make it simple

Can we get in a msgbox only the brackets

i.e
INPUT=(abc(xyz(123)ccc)((456)789)
OUTPUT=((())(())

Everything that is not "(" or is not ")" should be delected

Originally Posted by mikerickson

Another function you might use is

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FYI, there are two typos (highlighted).

Hi jindon
seems it will be helpful.but I am testing the likely combinations and let you know the outcome.

Thanks mikericson and jindo

and thanks xlnitwit for the spotting the typos in mike's solution.

Cant this just be done as a formula?

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