[2008] Show Windows "Open With" dialog when no file association found

Hi,

I am using Process.Start(fileName) to let Windows open certain files outside of my application with their 'default application'.

If "fileName" for example is a ".txt" file, it opens the file in Notepad. If it is a folder, it opens windows explorer with that folder selected.

This all seems to work well, except when the user enters a file that has no application 'linked' to it... (a file that has no file associations)

I have wrapped the Process.Start() in a Try/Catch block so that any errors are caught and displayed.
When I now open a file without any file associations (windows does not know with which application to open it) it tells me that it cannot find any application to open it with. (The exact error is in dutch so I don't think that's going to be much help here, but I think it's clear what's happening, no?)


What I would like to happen in this case is that the windows "Open With" dialog comes up, asking the user to select an application to open the file with.
Now it just displays the error message and continues, making it impossible for the user to open the file unless he browses to the file manually and sets up the file association, something many users probably don't know how to do...


So is this possible?

Try this:

Code:

---

Private Sub OpenWith(ByVal filePath As String)
        Dim psi As New ProcessStartInfo("rundll32.exe")
        With psi
            .Arguments = String.Format("shell32.dll,OpenAs_RunDLL {0}", filePath)
        End With
        Process.Start(psi)
    End Sub

---

To use it, you just call the sub and pass in the full path to the file you want to use "open with".

Thanks, it does work, but it doesn't do exactly what I want it to do unfortunately...

Now, it always seems to show the Open With dialog, even if there IS a file association. If there already is a file association, I want it to open with that default application and not ask the user.

I only want it to ask the user when windows cannot find any file association.


I suppose I could always just try to "Process.Start" the file normally, and try the "Open With" solution when it encounters an error, but that seems like a dirty workaround, there surely is a better way?

You just put Process.Start(filePath) in a try/catch block, and in the catch block, you use open with.

Code:

---

Private Sub OpenThisFile(ByVal filePath As String)
        If System.IO.File.Exists(filePath) Then
            Try
                Process.Start(filePath)
            Catch ex As Exception
                Dim psi As New ProcessStartInfo("rundll32.exe")
                psi.Arguments = String.Format("shell32.dll,OpenAs_RunDLL {0}", filePath)
                Process.Start(psi)
            End Try
        Else
            MessageBox.Show("File doesn't exist!")
        End If
    End Sub

---

Or you can look in the registry to see if the file type has an association.

Code:

---

        'HKEY_CLASSES_ROOT
        Dim HKCR As Microsoft.Win32.RegistryKey = Microsoft.Win32.Registry.ClassesRoot

        'extension to test for
        Dim FileType As Microsoft.Win32.RegistryKey = HKCR.OpenSubKey(".hght")

        If FileType IsNot Nothing Then
            MessageBox.Show("There is a 'default application'.")
        Else
            MessageBox.Show("No default application.")
        End If

---

stanav ... in your solution is there any way to skip the first screen (see below) & go straight to the second screen where you actually select the application you want to use?

.

I don't think there's a way to bypass that 1st dialog. That's how Windows works. If you try to open a file with unregistered file type manually by double-click on it, you'll see that 1st dialog. If you try right-click on the file, you won't see the option "Open With..." as with a registered file type.

I think I've seen some registry hack to turn off that first prompt but nobody reall likes programs changing global settings (esp. when unannounced). Also, to combine the given suggestions so far, there is a property in the ProcessStartInfo class that does what NickThissen wants (mentioned last here).

Code:

---

        Dim openWithPrInfo As New ProcessStartInfo

        openWithPrInfo.FileName = "c:\asdf.jjj"
        openWithPrInfo.ErrorDialog = True

        Process.Start(openWithPrInfo)

---

EDIT

Thanks Half, that did the trick perfectly!