Remove strings within brackets

'EmptyParentheses

' Desc: Accepts a string, then returns that string with any contents that appear within parentheses truncated.
' Function should return correct results for any # of Parens, but they must be all matched pairs AND in correct left/right order.

' Usage: EmptyParentheses("Original data - Remove (Sep 09-10) strings (Oct 09-10) within (Nov 09-10) brackets (Dec 09-10)")

' Input = "Original data - Remove (Sep 09-10) strings (Oct 09-10) within (Nov 09-10) brackets (Dec 09-10)"
' Output= "Original data - Remove () strings () within () brackets ()"

' Notes: This is a custom '#' IDEA function, used within the IDEA interface and stored in a file as "EmptyParentheses.ideafunc"

' Created: 2014/9/26
' Author: Rob Gentile
'
' Revisions: 2014/9/26 RMG Initial Script

Dim numLeftParens As Integer
Dim numRightParens As Integer
Dim numParensPairs As Integer
Dim ThisParensPair As Integer
Dim ParenthesesMatch As Boolean

'First, do the parentheses match? Return error if they do not.

numLeftParens = NumberOfCharsInString(TheString, "(")
numRightParens = NumberOfCharsInString(TheString, ")")
If numLeftParens = numRightParens Then
ParenthesesMatch = True
numParensPairs = numLeftParens
emptyparentheses = "Good = - the Parentheses are matched -- there are " & numLeftParens & "'" & "(" & "' , and there are " & numRightParens & " '" & ")" & "' ! "
Else
ParenthesesMatch = False
emptyparentheses = "Error - the Parentheses are not matched -- there are " & numLeftParens & "'" & "(" & "' , but there are " & numRightParens & " '" & ")" & "' ! "
End If

'So,OK, parentheses match, and we need to get the pieces we do want.

If ParenthesesMatch Then
emptyparentheses = "" 'reset variable
emptyparentheses = iSplit( TheString , "", "(", 1) +"(" 'grab initial text before first LEFT paren
For ThisParensPair = 1 To (numParensPairs - 1)
emptyparentheses = emptyparentheses +")" + iSplit( TheString , ")", "(", ThisParensPair) +"("
Next ProcessEachParensPairs
emptyparentheses = emptyparentheses + ")"
emptyparentheses = emptyparentheses + iSplit( TheString , "", ")", (ThisParensPair +1)) 'grab Last bit of text after last RIGHT paren
End If

'MsgBox EmptyParentheses

End Function

Function NumberOfCharsInString(TheString As String, TheChar As String) As Integer

Dim numLedftParens As Integer
Dim cnt As Integer
Dim StringLength As Integer
Dim CharInString As Integer
Dim ThisChar As String
Dim CharToCount As String

CharToCount = TheChar
cnt = 0
StringLength = Len(TheString)
For CharInString = 1 To StringLength
ThisChar = Mid( TheString, CharInString, 1)
If ThisChar = CharToCount Then cnt = cnt + 1
Next
NumberOfCharsInString = cnt

End Function