OpenFileDialog component in Visual Studio 2017 - visual-studio-2017

Question

I'm working on a VB project in Visual Studio 2017. It's a Blank App (Universal Windows) project. When trying to work with this type of app, it doesn't seem to have an OpenFileDialog like the Windows Forms App (.NET Framework) has. Is there a way to do one of two things:
Create a Windows Forms App that has the same look and feel as the Blank App (Universal Windows)
Add the OpenFileDialog option to the Blank App (Universal Windows)

Answer 1

The following code is a VB version of the MS example at https://learn.microsoft.com/en-us/uwp/api/Windows.Storage.Pickers.FileOpenPicker.
Imports Windows.Storage
Imports Windows.Storage.Pickers
Public NotInheritable Class MainPage
Inherits Page
Private Async Sub Button_Click(sender As Object, e As RoutedEventArgs)
Dim openPicker As New FileOpenPicker()
openPicker.ViewMode = PickerViewMode.Thumbnail
openPicker.SuggestedStartLocation = PickerLocationId.PicturesLibrary
openPicker.FileTypeFilter.Add(".jpg")
openPicker.FileTypeFilter.Add(".jpeg")
openPicker.FileTypeFilter.Add(".png")
Dim file As StorageFile = Await openPicker.PickSingleFileAsync()
If (file IsNot Nothing) Then
Debug.WriteLine($"Picked File: {file.Name}")
Else
Debug.WriteLine("Operation Cancelled.")
End If
End Sub
End Class

Answer 2

What I ended up doing to get the same(ish) look was to just play around with some of the properties of the form and the buttons. This gave me the look that I was after. It wasn't exact, but I'll take it.
As for OpenFileDialog, I ended up using the following:
Dim myStream As IO.Stream = Nothing
Dim openFileDialog1 As New OpenFileDialog()
' Open file dialog parameters
openFileDialog1.InitialDirectory = "c:\" ' Default open location
openFileDialog1.Filter = "Executable Files (*.exe)|*.exe|All Files (*.*)|*.*"
openFileDialog1.FilterIndex = 2
openFileDialog1.RestoreDirectory = True
If openFileDialog1.ShowDialog() = System.Windows.Forms.DialogResult.OK Then
Try
myStream = openFileDialog1.OpenFile()
If (myStream IsNot Nothing) Then
' Insert code to read the stream here.
Textbox1.Text = openFileDialog1.FileName
' Even though we're reading the entire path to the file, the file is going to be ignored and only the path will be saved.
' Mostly due to me lacking the ability to figure out how to open just the directory instead of a file. Resolution threadbelow.
' http://www.vbforums.com/showthread.php?570294-RESOLVED-textbox-openfiledialog-folder
' Setting the public variable so it can be used later
Dim exepath As String
exepath = IO.Path.GetDirectoryName(Me.txtExeLocation.Text)
End If
Catch Ex As Exception
MessageBox.Show("Cannot read file from disk. Original error: " & Ex.Message)
Finally
' Check this again, since we need to make sure we didn't throw an exception on open.
If (myStream IsNot Nothing) Then
myStream.Close()
End If
End Try
End If