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我在VB中编写如下打开2004图形文件的代码
Option Explicit
Public AcadAppObj As AcadApplication
Public Curdwg As AcadDocument
Private Sub Command1_Click()
'连接Autocad2004
On Error Resume Next
Set AcadAppObj = GetObject(, "autocad.application.16")
If Err Then
Err.Clear
Set AcadAppObj = CreateObject("autocad.application.16")
If Err Then
MsgBox "不能运行AutoCAD!"
Exit Sub
End If
End If
AcadAppObj.Visible = True
Set Curdwg = AcadAppObj.ActiveDocument
Curdwg.Open "E:\习题\平面图\齿轮.dwg"
Curdwg.Active
End Sub
为何说 Curdwg.Active 属性的无效使用,AcadAppObj和Curdwg的值溢出上下文,且autocad2004并没有打开所要打开的图形文件,请问各位究竟是何原因。 |
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