Start a document based on its filename

Author:

Steve McMahon(steve@vbaccelerator.com)

Keywords:

API,Shell

Updated:

01/08/98

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Submit

Starting any document file based on its file name only is very simple in Windows 9x and Windows NT using the ShellExecute function. This tip shows how simple it is - you only really need one declare and one line of code!

Start a new project in VB. Add a Command button to the project's form, then add the following code:

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 Sub Command1_Click()
Dim lR As Long
Dim sFile As String
Dim iFile As Integer

    ' Create a text file to test:
    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, , "This is a text file used for testing the ShellExecute API function."
    Close #iFile

    ' Start the text file using its filename. Windows will
    ' check what exe is associated with .TXT files (by default
    ' this is Notepad) and start the application with the file
    ' opened:
    lR = ShellExecute(Me.hWnd, "Open", sFile, "", "", vbNormalFocus)
    If (lR < 0) Or (lR > 32) Then
        ' success
    Else
        MsgBox "Failed to start '" & sFile & "'", vbInformation
    End If
End Sub

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