Open Photographs in an Access Form

Hello

I am Abdal and I would be glad to help you with your question.

Yes, it is possible to create a command button on your List Form that opens the photo based on the digital file name for the selected record. You can use VBA (Visual Basic for Applications) code to accomplish this. The code would use the FileSystemObject to open the file and the value of the "Digital File Name" field to construct the file path. You can then assign the code to the On Click event of the command button. Here is an example of code that you could use to open the file:

Private Sub cmdOpenFile_Click() Dim fso As Object Dim filePath As String

' Get the value of the "Digital File Name" field filePath = Me! [Digital File Name]

' Construct the full file path filePath = "C:\Photos" & filePath

' Create a FileSystemObject Set fso = CreateObject("Scripting.FileSystemObject")

' Open the file fso. GetFile(filePath). Open End Sub

You would need to replace 'Digital File Name' with the name of your field and the 'C:\Photos' with the path of your photos directory.

I hope this information helps.

Regards,

Abdal

Sure, it real easy. I'm assuming here that the Digital File Name includes the full path and filename. So what I would do is use the Double click event of the control bound that field and use this code:

Application.FollowHyperlink Me.txtFilename

Where txtFilename is the name of the control bound to the field. This will open the fiel in your default photo editor.

If you only store the filename and not the full path, you sould need to do something like:

Application.FollowHyperlink "C:\photos" & Me.txtFilename

Alternatively, you could simply add an unbound image control to the form and set the Controlsource to the field.

Abdal,

Thank you for your reply. I tried the code you suggested and it would not work for me. When I clicked my Open Image command button nothing happened. Following is the code I wrote as copied from the Code bulider page of the Open Image command button:

Private Sub cmdOpenFile_Click()

Dim fso As Object 

Dim filePath As String 

'Get the value of the "Digital File Name" field

filePath = Me![Digital File Name] 

' Construct the full file Path

filePath = "D:\Documents\Family Tree Maker\Wright-Jagoe(1) Media" & filePath 

' Create a FileSystemObject

Set fso = CreateObject("Scripting.FileSystemObject') 

' Open the file

fso.GetFile(filePath).Open 

End Sub

I must have made some mistake as I am not to far advanced in Access, although I do follow directions pretty well.

Hope you can help - Walt

Did you try the suggestion I made?

While Abdal's solution can work, it is using a cannon to swat a fly.

Hello

Thanks for the update.

You may want to include error handling in case the file is not found or the path is incorrect. You can use the On Error Resume Next statement to catch any errors and then use the On Error GoTo statement to handle them.

For example:

Private Sub cmdOpenFile_Click() On Error Resume Next Dim fso As Object Dim filePath As String

' Get the value of the "Digital File Name" field filePath = Me! [Digital File Name]

' Construct the full file path filePath = "C:\Photos" & filePath

' Create a FileSystemObject Set fso = CreateObject("Scripting.FileSystemObject")

' Open the file fso. GetFile(filePath). Open

' Error handling If Err.Number <> 0 Then MsgBox "Error opening file. Please check the file name and path.", vbExclamation Exit Sub End If End Sub

This will display a message box if an error occurs while trying to open the file, allowing the user to check the file name and path.

It is also recommended to add a button caption to the button in the design view so the user knows what the button does.

Lastly, you can also add a filter to the file path if the user has a different file type other than the default one. This will ensure that the file opens in the correct application.

For example:

filePath = "C:\Photos" & filePath & ".jpg"

This will add .jpg to the file path, ensuring that the file opens in the correct application for jpeg files.

While the code you gave seems to have some typo error.

On this line:

Set fso = CreateObject("Scripting.FileSystemObject')

There is an extra single quote at the end of the string that should be removed. It should be:

Set fso = CreateObject("Scripting.FileSystemObject")

Also, please make sure that the directory "D:\Documents\Family Tree Maker\Wright-Jagoe(1) Media" exists and that the file name you have in the "Digital File Name" field matches the file name in the directory.

Another thing that you can do is to add a MsgBox command before the fso. GetFile(filePath). Open command, so that you can check the value of the filePath variable. This will help you to understand if the path is correct or not.

For example:

MsgBox filePath

This will display the filePath value in a message box, so you can see if it is what you expect.

Finally, you can also include error handling as I mentioned before, so that if the file is not found or the path is incorrect, a message box will be displayed, allowing the user to check the file name and path.

I hope this information helps.

Regards,

Abdal