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Launch the File Properties dialog box for a given file

by Anonymous (267 Submissions)
Category: Files/File Controls/Input/Output
Compatability: Visual Basic 4.0 (32-bit)
Difficulty: Unknown Difficulty
Originally Published: Wed 1st December 1999
Date Added: Mon 8th February 2021
Rating: (1 Votes)

Launch the File Properties dialog box for a given file

API Declarations


cbSize As Long
fMask As Long
hwnd As Long
lpVerb As String
lpFile As String
lpParameters As String
lpDirectory As String
nShow As Long
hInstApp As Long
lpIDList As Long
lpClass As String
hkeyClass As Long
dwHotKey As Long
hIcon As Long
hProcess As Long
End Type

Public Const SEE_MASK_INVOKEIDLIST = &HC
Public Const SEE_MASK_NOCLOSEPROCESS = &H40
Public Const SEE_MASK_FLAG_NO_UI = &H400

Declare Function ShellExecuteEX Lib "shell32.dll" Alias "ShellExecuteEx" (SEI As SHELLEXECUTEINFO) As Long



Rate Launch the File Properties dialog box for a given file




    Dim SEI As SHELLEXECUTEINFO
    Dim lngReturn As Long
     
    With SEI
        .cbSize = Len(SEI)
        .fMask = SEE_MASK_NOCLOSEPROCESS Or _
         SEE_MASK_INVOKEIDLIST Or SEE_MASK_FLAG_NO_UI
        .hwnd = OwnerhWnd
        .lpVerb = "properties"
        .lpFile = FileName
        .lpParameters = vbNullChar
        .lpDirectory = vbNullChar
        .nShow = 0
        .hInstApp = 0
        .lpIDList = 0
    End With
     
    lngReturn = ShellExecuteEX(SEI)
    
End Sub

'for example, to show file properties dialog box for autoexec.bat file, use...
Call ShowProps("c:\autoexec.bat", Me.hwnd)

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