Question

My WPF application gets a file from the user with Microsoft.Win32.OpenFileDialog()…

Private Sub ButtonUpload_Click(...)     Dim FileOpenStream As Stream = Nothing     Dim FileBox As New Microsoft.Win32.OpenFileDialog()     FileBox.InitialDirectory = Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments)     FileBox.Filter = 'Pictures (*.jpg;*.jpeg;*.gif;*.png)|*.jpg;*.jpeg;*.gif;*.png|' & _                      'Documents (*.pdf;*.doc;*.docx;)|*.pdf;*.doc;*.docx;|' & _                      'All Files (*.*)|*.*'     FileBox.FilterIndex = 1     FileBox.Multiselect = False     Dim FileSelected As Nullable(Of Boolean) = FileBox.ShowDialog(Me)     If FileSelected IsNot Nothing AndAlso FileSelected.Value = True Then         Try             FileOpenStream = FileBox.OpenFile()             If (FileOpenStream IsNot Nothing) Then                 Dim ByteArray As Byte()                 Using br As New BinaryReader(FileOpenStream)                     ByteArray = br.ReadBytes(FileOpenStream.Length)                 End Using                 Dim z As New ZackFile                 z.Id = Guid.NewGuid                 z.FileData = ByteArray                 z.FileSize = CInt(ByteArray.Length)                 z.FileName = FileBox.FileName.Split('\').Last                 z.DateAdded = Now                 db.AddToZackFile(z)                 db.SaveChanges()             End If         Catch Ex As Exception             MessageBox.Show('Cannot read file from disk. ' & Ex.Message, 'Fail', MessageBoxButton.OK, MessageBoxImage.Error)         Finally             If (FileOpenStream IsNot Nothing) Then                 FileOpenStream.Close()             End If         End Try     End If End Sub 

And my ASP.NET MVC application serves it up for download at a web site with FileStreamResult()…

Public Class ZackFileController     Inherits System.Web.Mvc.Controller      Function Display(ByVal id As Guid) As FileStreamResult         Dim db As New EfrDotOrgEntities         Dim Model As ZackFile = (From z As ZackFile In db.ZackFile _                                 Where z.Id = id _                                 Select z).First         Dim ByteArray As Byte() = Model.ImageData         Dim FileStream As System.IO.MemoryStream = New System.IO.MemoryStream(ByteArray)         Dim ContentType As String = ?????         Dim f As New FileStreamResult(FileStream, ContentType)         f.FileDownloadName = Model.FileName         Return f     End Function  End Class 

But FileStreamResult() needs a content type string. How do I know the correct content type of my file?

Answer 1

I’ve replaced the FileExtension column in my database table with a ContentType column.

I populate it when I upload a file.

Private Sub ButtonUpload_Click(...)     ...     Dim FileExtension As String = '.' + FileBox.FileName.Split('.').Last.ToLower     z.ContentType = ContentType(FileExtension)     ... End Sub 

I determine the content type with this function:

Function ContentType(ByVal FileExtension As String) As String     Dim d As New Dictionary(Of String, String)     'Images'     d.Add('.bmp', 'image/bmp')     d.Add('.gif', 'image/gif')     d.Add('.jpeg', 'image/jpeg')     d.Add('.jpg', 'image/jpeg')     d.Add('.png', 'image/png')     d.Add('.tif', 'image/tiff')     d.Add('.tiff', 'image/tiff')     'Documents'     d.Add('.doc', 'application/msword')     d.Add('.docx', 'application/vnd.openxmlformats-officedocument.wordprocessingml.document')     d.Add('.pdf', 'application/pdf')     'Slideshows'     d.Add('.ppt', 'application/vnd.ms-powerpoint')     d.Add('.pptx', 'application/vnd.openxmlformats-officedocument.presentationml.presentation')     'Data'     d.Add('.xlsx', 'application/vnd.openxmlformats-officedocument.spreadsheetml.sheet')     d.Add('.xls', 'application/vnd.ms-excel')     d.Add('.csv', 'text/csv')     d.Add('.xml', 'text/xml')     d.Add('.txt', 'text/plain')     'Compressed Folders'     d.Add('.zip', 'application/zip')     'Audio'     d.Add('.ogg', 'application/ogg')     d.Add('.mp3', 'audio/mpeg')     d.Add('.wma', 'audio/x-ms-wma')     d.Add('.wav', 'audio/x-wav')     'Video'     d.Add('.wmv', 'audio/x-ms-wmv')     d.Add('.swf', 'application/x-shockwave-flash')     d.Add('.avi', 'video/avi')     d.Add('.mp4', 'video/mp4')     d.Add('.mpeg', 'video/mpeg')     d.Add('.mpg', 'video/mpeg')     d.Add('.qt', 'video/quicktime')     Return d(FileExtension) End Function 

This works, but it seems inelegant.