我一直在使用以下宏將括號中的專案提取到 word 中的注釋中：
'
' CommentBubble Macro
'
'
Dim myRange As Range
Set myRange = ActiveDocument.Content
searchtext = "\(*\)"
With myRange.Find
.MatchWildcards = True
Do While .Execute(findText:=searchtext, Forward:=True) = True
If Len(myRange.Text) > 4 Then
ActiveDocument.Comments.Add myRange, myRange.Text
myRange.Text = ""
End If
Loop
End With
End Sub
我將文本長度設定為 > 4 的原因是因為這些是法律檔案，我不想在以下條件下隔離具有諸如“的字串：(i) 條件 1，(ii) 條件 2 等.”
但是，這是上面代碼中斷的一段文本：
This is sample text (with some additional text) that does stuff (with more stuff) and represents 39.4% of shares on the effective date (before giving effect, with some conditions such as ( some stuff (i) and some stuff (ii) with final stuff) and more final stuff) which is subject to (some conditions here) and conclude here.
如果你運行這個，你會得到以下結果：
This is sample text that does stuff and represents 39.4% of shares on the effective date and some stuff (ii) with final stuff) and more final stuff) which is subject to and conclude here.
如您所見，嵌套括號會引起一些麻煩。有什么建議嗎？
謝謝！
uj5u.com熱心網友回復：
您正在嘗試匹配括號，這在 Word 中是一項困難且吃力不討好的任務，因為 Word 僅將左括號和右括號視為單個字符，而不會自動按單詞進行匹配。下面的代碼查找匹配的括號，消除尾隨空格，處理沒有括號的情況，如果有不平衡的錯誤，則會出錯。我已經留在除錯陳述句中，以便您可以取消注釋它們以查看發生了什么。
Option Explicit
Public Sub ttest()
Dim myRange As Word.Range
Set myRange = ActiveDocument.StoryRanges(wdMainTextStory)
myRange.Collapse direction:=wdCollapseStart
Set myRange = NextParenRange(myRange)
Do Until myRange Is Nothing
DoEvents
Debug.Print myRange.Text
Dim myDupRange As Word.Range
Set myDupRange = myRange.Duplicate
myRange.Collapse direction:=wdCollapseEnd
If myDupRange.Characters.Last.Next.Text = " " Then myDupRange.MoveEnd Count:=1
myDupRange.Delete
Set myRange = NextParenRange(myRange)
Loop
End Sub
Public Function NextParenRange(ByVal ipRange As Word.Range) As Word.Range
Const OpenP As String = "("
Const CloseP As String = ")"
Dim myRange As Word.Range
Set myRange = ipRange.Duplicate
'If myRange.Start <> myRange.End Then myRange.Collapse direction:=wdCollapseStart
'exit if no parentheses exist
'Debug.Print myRange.Start
If myRange.MoveUntil(cset:=OpenP) = 0 Then
Set NextParenRange = Nothing
Exit Function
Else
'Debug.Print myRange.Start
Dim myParenCount As Long
myParenCount = 1
myRange.MoveEnd Count:=1
End If
Do Until myParenCount = 0
' allows VBA to respond to a break key press
DoEvents
' if we run out of parentheses before we get back to zero then flag an error
If myRange.MoveEndUntil(cset:=OpenP & CloseP) = 0 Then
VBA.Err.Raise 17, "Unbalanced parentheses in document"
End If
myRange.MoveEnd Count:=1
'Debug.Print myRange.Characters.Last.Text
'Debug.Print myRange.Characters.Last.Next.Text
myParenCount = myParenCount IIf(myRange.Characters.Last.Text = OpenP, 1, -1)
Loop
Set NextParenRange = myRange.Duplicate
End Function
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