如何在ACCESS窗体的里打开EXCEL文件

cangqiong

如何在ACCESS窗体的里打开EXCEL文件

我在窗体里做了一个command按钮....怎么现实这功能  

按钮一按打开D:\文件\文件.XLS

Grant

Access调用打开已关联的外部文件



Option Compare Database
    Private Declare Function ShellExecute Lib "shell32.dll" Alias "ShellExecuteA" _
     (ByVal hwnd As Long, ByVal lpszOp As String, _
     ByVal lpszFile As String, ByVal lpszParams As String, _
     ByVal LpszDir As String, ByVal FsShowCmd As Long) _
     As Long
    Const SW_SHOW = 1
     
Private Sub Enter_Click()
ShellExecute 0, "Open", "c:\aa.doc", "", "", SW_SHOW
End Sub

glw

在别的论坛上看到的，有三种方法



Private Sub command2_Click()
Dim ex As New Excel.Application
Set ex = CreateObject("excel.application")
Set exwbook = ex.Workbooks.Open(CurrentProject.Path & "\test.xls")
ex.Visible = True
End Sub

Private Sub command3_Click()
Dim exfile As String, shellstg As String
exfile = CurrentProject.Path & "\test.xls"

'注意下面代码的引号用法
   shellstg = """D:\Program Files\Microsoft Office\Office\EXCEL.EXE"" """ & exfile & """"
Call Shell(shellstg, 1)


End Sub

Private Sub Form_Load()
'设置command1的HyperlinkAddress属性，即超级链接属性
  Command1.HyperlinkAddress = CurrentProject.Path & "\test.xls"
End Sub

cangqiong

谢谢