Open Photographs in an Access Form

Anonymous
2023-01-26T23:26:10+00:00

I have a datatbase of photograph file names. In the DB I have a List Form that has a "Digital File Name" field. Question: Is it possable to create a command button on my List Form that will open the photo (that is stored in my Photos File on my C: drive) based on the digital file name for the selected record?

I am using Access 2016

Thanks Walt

Microsoft 365 and Office | Access | For home | Other

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  1. ScottGem 68,840 Reputation points Volunteer Moderator
    2023-03-13T23:53:41+00:00

    You need to use DIR to check if the file exists

    If DIR(Me.DigitalFilename) = "" Then

    MsgBox "File doesn't exist"

    Else

    Application.Followhyperlink Me.DigitalFilename

    End If

    Make sure you don't use both the Click and double click events, just us the Click event.

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  2. Anonymous
    2023-03-14T19:56:42+00:00

    Scott,

    I entered the following code:

    Private Sub OpenImageBtn_Click(Cancel As Integer)

    Application.FollowHyperlink Me.DigitalFileName

    If Dir(Me.DigitalFileName) = "" Then 

    MsgBox "File doesn't exist"

    Else 

    Application.FollowHyperlink Me.DigitalFileName

    End If

    End Sub

    When I tried to open the search form I was asked for a Parameter. Also when in Code View, I tried to use the run button in the Debug Tool Bar and a popup box appeared asking for a Macro. I've done something wrong that i don't see.

    Walt

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  3. Anonymous
    2023-03-14T19:57:39+00:00

    Only the Click is used.

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  4. Anonymous
    2023-03-14T21:36:38+00:00

    Private Sub OpenImageBtn_Click(Cancel As Integer)

    Hi Walt,

    The Click event has no Cancel argument, so try:

    Private Sub OpenImageBtn_Click()

    It is the DblClick event that has the Cancel argument.

    Imb.

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  5. Anonymous
    2023-04-21T21:22:51+00:00

    I am trying to include an error trap with my "Open Image Button" so that when searched and no digital image file is found a message box will come up with "File does not exist".

    The following is the code I have entered:

    Private Sub OpenImageBtn_Click()

    Application.FollowHyperlink Me.DigitalFileName

    If Dir(Me.DigitalFileName) = "" Then

    MsgBox "File doesn't exist"

    Else

    Application.FollowHyperlink Me.DigitalFileName

    End If

    End Sub

    When I click the "Open Image Button" the message box don't appear and the image is opened twice. Can anyone tell me what's wrong with this code.

    Thanks, Walt

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