fileupload

DrFunk · June 24th, 2010, 03:18 AM

Hello,

Due to some hostingrestrictions I am obliged to use access as a database.

Everything goes well untill I get to the fileupload in chapter 13.

Apparantely a accessdatasource does not work with e.newobject very well.

Ctype(e.NewObject, Picture) is not a member of SqlDataSourceCommandEventArgs...

Code:

 Protected Sub AccessDataSource2_Inserting(ByVal sender As Object, ByVal e As System.Web.UI.WebControls.SqlDataSourceCommandEventArgs) Handles AccessDataSource2.Inserting
        Dim mypicture As Picture = CType(e.NewObject, Picture)

        Dim FileUpload1 As FileUpload = _
      CType(ListView1.InsertItem.FindControl("FileUpload1"), FileUpload)
        Dim virtualFolder As String = "~/pics/"
        Dim physicalFolder As String = Server.MapPath(virtualFolder)
        Dim fileName As String = Guid.NewGuid().ToString()
        Dim extension As String = System.IO.Path.GetExtension(FileUpload1.FileName)

        FileUpload1.SaveAs(System.IO.Path.Combine(physicalFolder, fileName + extension))
        mypicture.ImageUrl = virtualFolder + fileName + extension

    End Sub

Help is very much appriciated

Imar · June 24th, 2010, 04:47 AM

Hi there,

The AccessDataSource is not strongly typed and doesn't provide access to an object such as a Picture. Instead, you need to access its Parameters collection. Something like this should work:

Code:

e.Command.Parameters["ImageUrl"].Value = virtualFolder + fileName + extension;

Hope this helps,

Imar