In this article I will explain with an example, how to convert HttpPostedFile class object to Byte Array using C# and VB.Net.
The HttpPostedFile class object holds the File data for the ASP.Net FileUpload control.
In order to get the Byte Array from the HttpPostedFile class object, ReadBytes method of the BinaryReader class is used.
HTML Markup
The following HTML Markup consists of an ASP.Net FileUpload control, a Button and an Image control.
<asp:FileUpload ID="FileUpload1" runat="server" />
<asp:Button ID="btnUpload" runat="server" Text="Upload"
onclick="btnUpload_Click" />
<hr />
<asp:Image ID="Image1" Visible = "false" runat="server" Height = "100" Width = "100" />
Namespaces
You will need to import the following namespace.
C#
VB.Net
Convert HttpPostedFile to Byte Array using C# and VB.Net
When the Upload button is clicked, the Image file is read into a Byte Array using the BinaryReader class object.
The Byte Array is then saved to a folder as Image file using the WriteAllBytes method of the File class.
Finally the Base64 encoded string is displayed on web page as Image using Image control.
C#
protected void btnUpload_Click(object sender, EventArgs e)
{
//Read the uploaded File as Byte Array from FileUpload control.
Stream fs = FileUpload1.PostedFile.InputStream;
BinaryReader br = new BinaryReader(fs);
byte[] bytes = br.ReadBytes((Int32)fs.Length);
//Save the Byte Array as File.
string filePath = "~/Files/" + Path.GetFileName(FileUpload1.FileName);
File.WriteAllBytes(Server.MapPath(filePath), bytes);
//Display the Image File.
Image1.ImageUrl = filePath;
Image1.Visible = true;
}
VB.Net
Protected Sub btnUpload_Click(sender As Object, e As EventArgs)
'Read the uploaded File as Byte Array from FileUpload control.
Dim fs As Stream = FileUpload1.PostedFile.InputStream
Dim br As New BinaryReader(fs)
Dim bytes As Byte() = br.ReadBytes(Convert.ToInt32(fs.Length))
'Save the Byte Array as File.
Dim filePath As String = "~/Files/" + Path.GetFileName(FileUpload1.FileName)
File.WriteAllBytes(Server.MapPath(filePath), bytes)
'Display the Image File.
Image1.ImageUrl = filePath
Image1.Visible = True
End Sub
Demo
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