Image File

Question

Hi

I would like to have any function or routine to open an image file located outside the access database without inserting the image to access form or report.

Best Regards

Answer 1

https://www.devhut.net/ms-access-vba-open-a-file/

Answer 2

Use an API call.

This is the complete code you can save as a module named, say, ShowDocumentFile:

Option Compare Database Option Explicit

' API declarations for OpenDocumentFile.
' Documentation:
'   https://docs.microsoft.com/en-us/windows/win32/api/shellapi/nf-shellapi-shellexecutea
'
Private Declare PtrSafe 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 Declare PtrSafe Function GetDesktopWindow Lib "USER32" () _
    As Long
    
' Enum for ShowWindow constants (selection).
' Documentation:
'   https://docs.microsoft.com/en-us/windows/win32/api/winuser/nf-winuser-showwindow
'
Public Enum SwShowCommand
    swShowNormal = 1
    swShowMinimized = 2
    swShowMaximized = 3
End Enum
   
' Open a document file using its default viewer application.
' Optionally, the document can be opened minimised or maximised.
'
' Returns True if success, False if not.
' Will not raise an error if the path or file is not found.
'
' 2022-03-12. Gustav Brock, Cactus Data ApS, CPH.
'
Public Function OpenDocumentFile( _
    ByVal File As String, _
    Optional ByVal ShowCommand As SwShowCommand = swShowNormal) _
    As Boolean

    Const OperationOpen     As String = "open"
    Const MinimumSuccess    As Long = 32
    ' Shall not have a value for opening a document.
    Const Parameters        As String = ""
    
    Dim Handle      As Long
    Dim Directory   As String
    Dim Instance    As Long
    Dim Success     As Boolean
    
    Handle = GetDesktopWindow
    Directory = Environ("Temp")

    Instance = ShellExecute(Handle, OperationOpen, File, Parameters, Directory, ShowCommand)
    ' If the function succeeds, it returns a value greater than MinimumSuccess.
    Success = (Instance > MinimumSuccess)
    
    OpenDocumentFile = Success

End Function

Answer 3

Hi. Check out the Application.FollowHyperlink method. Also, look into the ShellExecute API.