String VBA-从GUID中删除括号{}

您好，我刚刚实施了新的修复程序，用于在microsoft更新后生成GUID代码。我是VBA新手，你能帮我从下面的代码中删除用GUID生成的{}花括号吗

 Private Type GUID_TYPE
            Data1 As Long
            Data2 As Integer
            Data3 As Integer
            Data4(7) As Byte
End Type
Private Declare PtrSafe Function CoCreateGuid Lib "ole32.dll" (guid As GUID_TYPE) As LongPtr
Private Declare PtrSafe Function StringFromGUID2 Lib "ole32.dll" (guid As GUID_TYPE, ByVal lpStrGuid As LongPtr, ByVal cbMax As Long) As LongPtr
Public Function GetGUID()
            Dim guid As GUID_TYPE
            Dim strGuid As String
            Dim retValue As LongPtr
            Const guidLength As Long = 39 'registry GUID format with null terminator {xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx}

            retValue = CoCreateGuid(guid)
            If retValue = 0 Then
                            strGuid = String$(guidLength, vbNullChar)
retValue = StringFromGUID2(guid, StrPtr(strGuid), guidLength)
                            If retValue = guidLength Then

GetGUID = strGuid
                            End If
            End If

结束函数编辑了这一行

---

GetGUID=Replace（Replace（strGuid，“{，”），“}，”）

只需使用

Mid

删除第一个和最后一个字符。 您还可以使用字节数组来保存guid和结果字符串

要生成GUID，请执行以下操作：

Private Declare PtrSafe Function CoCreateGuid Lib "ole32.dll" (pguid As Byte) As Long
Private Declare PtrSafe Function StringFromGUID2 Lib "ole32.dll" (rguid As Byte, lpsz As Byte, ByVal cchMax As Long) As Long


Public Function GenerateGUID() As String
  Dim pguid(0 To 15) As Byte, lpsz(0 To 77) As Byte

  CoCreateGuid pguid(0)
  StringFromGUID2 pguid(0), lpsz(0), 39
  GenerateGUID = Mid$(lpsz, 2, 36)
End Function

---

例如：guid='{abcd}'；然后使用“new_guid=Replace（Replace（guid），{，，，}，”）