Access - query to delete part of a name field in parantheses

Please advise if there is a way to remove part of a field that is enclosed in parantheses or quotes

Example

"Scott, Thomas (Tom)" - becomes "Scott, Thomas"
"Values above average (high)" - becomes "Values above average"

I am able to use Like"*(*" to update or delete the entire record, but not modify just part of the field.

Thank you.
exp vgAsked:
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Ryan ChongBusiness Systems Analyst , ex-Senior Application EngineerCommented:
in your Update SQL statement, you can call a user-defined function... like:

in a Module, try add:
Public Function SField(ByVal f As String, ByVal Delimiter As String) As String
       On Error Resume Next
       SField = split(f, Delimiter)(0)
       if Err > 0 Then
           SField = f
      End If 
End Function

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then to run your Update SQL statement, try:
update yourTable set fieldname = SField(fieldname, ' (' )

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chaauCommented:
I think you do not need a VBA function for a simple query like this. This straightforward query will do:
Update Table1
SET column1 = Mid(column1,1,InStr(column1,'(')-1) 
WHERE column1LIKE '*(*'

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Dale FyeOwner, Developing Solutions LLCCommented:
Both of the previous responses assume that the text wrapped in parenthesis is at the end of your field, but you did not explicitly state that that was the case. So, I would agree with Ryan that a function would be your best approach.

Public Function fnStripParen(Byval SomeValue as Variant) as Variant

    Dim intLeftParen as integer, intRightParen as integer
    Dim strLeft as string, strRight as string

    if isnull(SomeValue) then Exit Function

    'default return value
    fnStripParen = SomeValue

    'No left paren
    intleftParen = instr(SomeValue, "(")
    if intLeftParen = 0 then exit function

    'No right Paren
    intRightParen = instr(intLeftParen, SomeValue, ")")
    if intRightParen = 0 then exit function

    strLeft = IIF(IntLeftParen = 1, "", Left(SomeValue, intLeftParen-1))
    strRight = iif(intRightParen = Len(SomeValue), "", Mid(SomeValue, intRightParen + 1))
    fnStripParen = strLeft & strRight

End Function

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exp vgAuthor Commented:
Thank you everyone
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Microsoft Access

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