Open Photographs in an Access Form

Here's what I tried that has something I did wrong as it doesn't work.

Private Sub OpenImageBtn_DblClick(Cancel As Integer)

Dim strFile As String

strFile1 = "D:\Documents\Family Tree Maker\Wright-Jagoe Media" & Me.DigitFileName1

strFile2 = "D:\Documents\Family Tree Maker\Laidlaw-Fowler Media" & Me.DigitFileName2

If Dir(strFile1) = "" Or Dir(strFile2) = "" Then

MsgBox "File does not exist!"

Else

Application.FollowHyperlink strFile

End If

End Sub

When you say something doesn't work, you need to explain why and how it doesn't work. But I do see the problem. Look at the code. At the end you are trying to open strFile. Do you define strFile anyplace?

So, lets backtrack a bit. This was your initial request; "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?"

So, some questions

In the record it appears you just store the filename and not the full path? Why?

Are the 2 folders the only ones you store media in?

Could the image file exist in both folders?

Is there something else in the record that might indicate what folder the file is in?

Why would a filename be added to a record when the file is not in a folder?

Answers:

There is no real reason that stored only the file name and not the entire path. I see where you're going. If I stored the path then there is no requirement for more code as the search would go to the appropriate folder automatically. however, for several thousand images, it's alot more typing.

These are the only two folders.

An image cannot be in both folders.

there is nothing else in the record that would indicate which folder contains the image.

There is no reason for a file name to be in a record if the image is not in either folder.

Walt

An image cannot be in both folders.

there is nothing else in the record that would indicate which folder contains the image.

There is no reason for a file name to be in a record if the image is not in either folder.

As the file can only be in one of two folders it is easy to return the correct full path with an expression like that below, illustrated here in the immediate window on my system:

FileName = TestBed.accdb

? IIf(Dir("C:\Users\kenws\Documents\Databases\TestBed" & FileName)<>"","C:\Users\kenws\Documents\Databases\TestBed" & FileName,"C:\Users\kenws\Documents\Databases\SomeOtherFolder" & FileName)

Which correctly returns:

C:\Users\kenws\Documents\Databases\TestBed\Testbed.accdb

Whereas with:

FileName = "SomeOtherFile.accdb"

The same expression correctly returns:

C:\Users\kenws\Documents\Databases\SomeOtherFolder\SomeOtherFile.accdb

The expression should only be called If Not IsNull(FileName) of course to allow for records with no FileName value.

I have tried to go back and figure out how the code shold be written and this is what I cam up with using the entire code and only changing the lines that were modified.

 Private Sub OpenImageBtn\_DblClick(Cancel As Integer)

 Dim strFile as String

 strFile1 = "D:\Documents\Family Tree Maker\Wright-Jagoe Media\" & Me.DigitFileName1

 strFile2 = "D:\Documents\Family Tree Maker\Laidlaw-Fowler Media\" & Me.DigitFileName2

 If DIR(strFile1) = "" OR DIR(strFile2) = "" Then

 MsgBox "File does not exist!"

 Else

 Application.FollowHyperlink strFile

 End If

 End Sub

I changed the path in strFile2 to reflect the second file where images are stored.

When I tried to use the Open Image Button, I got the following error message:

 Compile error:

 Method or Datat member not found