要以与文件后缀名关联的程序打开文档,在Windows 9X和Windows NT下可以用ShellExcute函数方便地实现。这则小技巧展示了会有多方便—你只需要一个声明和一行代码!
开始一个新项目。在Form上放一个Command button,然后加入以下代码:
Private Declare Function ShellExecute Lib "shell32.dll" Alias "ShellExecuteA" _
(ByVal hWnd As Long, ByVal lpOperation As String, _
ByVal lpFile As String, ByVal lpParameters As String, _
ByVal lpDirectory As String, ByVal nShowCmd As Long) As Long
Private Const SW_NORMAL=1 '(这些API常量可以用VB常量代替,比如vbNormalFocus)
Private Const SW_MAXIMIZE=3
Private Const SW_MINIMIZE=6
Private Const SW_SHOW = 5
Private Sub Command1_Click()
Dim lR As Long
Dim sFile As String
Dim iFile As Integer
' 创建一个测试用的文本文件
sFile = App.Path & "\SHELLTST.TXT"
On Error Resume Next
Kill sFile
On Error GoTo 0
iFile = FreeFile
Open sFile For Binary Access Write As #iFile
Put #iFile, , "这是一个测试文件,演示ShellExecute API函数。"
Close #iFile
' 依照文件名打开这个文本。Windows将会检查哪个可执行程序与.TXT关联
' (默认一般是Notepad),并运行程序打开文档
lR = ShellExecute(Me.hWnd, "Open", sFile, "", "", vbNormalFocus)
If (lR < 0) Or (lR > 32) Then
' 成功
Else
MsgBox "无法打开 '" & sFile & "'", vbInformation
End If
End Sub
当你点击command button,将会在项目所在目录创建一个文本文件,并用默认程序打开(一般是Notepad)。