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Home/ Questions/Q 890879
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T13:48:13+00:00 2026-05-15T13:48:13+00:00

I have an Image object. I have the Picture type set to linked, so

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I have an Image object.
I have the Picture type set to linked, so I can change the picture if I want.
I have the Picture property set to the picture name.

I would think that access would use relitive addressing and simple looking in the current directory for the image. But it does not and I get an error telling me it cannot find the picture.

Anyone have a solution? (Other than setting the Picture type to embedded or using the full file address?)

Thanks!

Update:

Tried this:

Private Sub Form_Load()
Dim file As String
file = CurrentDb().Name
file = Replace(file, ".mdb", ".bmp")
Me.Image46.Picture = file

End Sub

It works, except I still get the error message. I click O.K. and it works. Just need the error message to go away.

SOLUTION: Use the above code (or the code posted in the answer below) and then set the ‘picture type’ to “embedded” and then delete the ‘picture’ field so that it says “(none)”.
Save and run.
It should work.

THANKS!

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  1. Editorial Team
    Editorial Team
    2026-05-15T13:48:13+00:00Added an answer on May 15, 2026 at 1:48 pm

    You could set the property on the forms OnLoad event like this

    Me.imgMy_image.picture=getDBPath & “mypicture.bmp”
    

    Here is the getDBPath function

    Public Function GetDBPath() As String
        Dim strFullPath As String
        Dim I As Integer
    
        strFullPath = CurrentDb().Name
    
        For I = Len(strFullPath) To 1 Step -1
            If Mid(strFullPath, I, 1) = "\" Then
                GetDBPath = Left(strFullPath, I)
                Exit For
            End If
        Next
    End Function
    

    Before anyone comments yes I know in access 2000 and above you can use currentproject.path but I’m stuck in the land that time forgot so need that custom function, it still works with later versions of access

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