"Cor Ligthert" <notfirstname@planet.nl> wrote in message
news:uW8Q3M3JEHA.3216@tk2msftngp13.phx.gbl...
> Hi Brian,
>
> I do not know if I am the only one who does not understand it.
> Your talking about executing files in a database.
> I thought you only could retrieve a file from a database.
>
> However maybe I do not know something about processing a datatype in a
> database, I am curious to know which datatype I than have to use.
>
> Can you explain it a little bit more?
>
> Cor
>
>

> I have been working executeing files a user selects in a database, my
> problem is that the file type may or may not be associated in the end users
> system. How can I go about, when the user executes a file, if the file does
> not have an associated application with it, show the window in windows that
> asks the user what they would like to open it with? (The select program from
> list window) thanks!

"Cor Ligthert" <notfirstname@planet.nl> wrote in message
news:u5RjSB6JEHA.1132@TK2MSFTNGP12.phx.gbl...
> Hi Brian,
>
> Did you ever saw this sample of me, with that there is no saving to disk
> needed?
> The XML file is just to imitate a dataset from a database or a datareader.
>
> Cor
>
>
> Private abyt() As Byte
> Private fo As New OpenFileDialog
> Private sf As New SaveFileDialog
> Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As
> System.EventArgs) Handles Button1.Click
> 'Reading a picture and put it in a bytearray
> If fo.ShowDialog = DialogResult.OK Then
> Dim fs As New IO.FileStream(fo.FileName, _
> IO.FileMode.Open)
> Dim br As New IO.BinaryReader(fs)
> abyt = br.ReadBytes(CInt(fs.Length))
> br.Close()
> 'just to show the sample without a fileread error
> Dim ms As New IO.MemoryStream(abyt)
> Me.PictureBox1.Image = Image.FromStream(ms)
> End If
> End Sub
> Private Sub Button2_Click(ByVal sender As System.Object, ByVal _
> e As System.EventArgs) Handles Button2.Click
> 'writing a picture from a bytearray
> If sf.ShowDialog = DialogResult.OK Then
> Dim fs As New IO.FileStream(sf.FileName, _
> IO.FileMode.CreateNew)
> Dim bw As New IO.BinaryWriter(fs)
> bw.Write(abyt)
> bw.Close()
> End If
> End Sub
> Private Sub Button3_Click(ByVal sender As System.Object, ByVal _
> e As System.EventArgs) Handles Button3.Click
> 'writing a bytearray to a dataset
> Dim ds As New DataSet
> ds.Tables.Add(New DataTable("Photo"))
> ds.Tables(0).Columns.Add(New DataColumn("Sample"))
> ds.Tables(0).Columns(0).DataType =
> System.Type.GetType("System.Byte[]")
> ds.Tables(0).Rows.Add(ds.Tables(0).NewRow)
> ds.Tables(0).Rows(0)(0) = abyt
> Dim sf As New SaveFileDialog
> If sf.ShowDialog = DialogResult.OK Then
> ds.WriteXml(sf.FileName, XmlWriteMode.WriteSchema)
> End If
> End Sub
> Private Sub Button4_Click(ByVal sender As System.Object, _
> ByVal e As System.EventArgs) Handles Button4.Click
> 'reading a picture from a dataset
> Dim ds As New DataSet
> If fo.ShowDialog = DialogResult.OK Then
> ds.ReadXml(fo.FileName)
> End If
> abyt = CType(ds.Tables(0).Rows(0)(0), Byte())
> Dim ms As New IO.MemoryStream(abyt)
> Me.PictureBox1.Image = Image.FromStream(ms)
> End Sub
> ///
>
>

> it's not a picture image, but an image data type in SQL server, the actual
> file could be anything from a picture or text document to a 100MB data
> file...