How to work with FileUpload Control

Question

Dear All I am working on FileUpload Control. By using this control when i select any image that image should be displayed in image button. How to achieve this can any one help me Thanks in advance

paul daniel · Answer

To Load the image in SQL Server (use this code) '#Region Reading file fs = New FileStream(Filename FileMode.Open) ' Finding out the size of the file to be uploaded Dim fi As FileInfo = New FileInfo(Filename) Dim temp As Long = fi.Length Dim lung As Integer = Convert.ToInt32(temp) ' Reading the content of the file into an array of bytes. Dim picture As Byte() = New Byte(lung - 1) {} fs.Read(picture 0 lung) fs.Close() conn = New SqlConnection(ConnectionString) Dim picstr As New SqlParameter picstr = insertStatement.Parameters.Add(@picstrText) Dim picstr1 As String picstr.Value = picture Dim l As String l = BitConverter.ToString(picture) insertStatement.CommandText = update organization set companyimage=@picstr where organization_id=3 insertStatement.CommandType = CommandType.Text l = Replace(l - ) 'insertStatement.CommandText = insert into imagetbl (imageschkimageId) values( & l & ) To Retrive the Image Dim DAdap As New SqlClient.SqlDataAdapter Dim DS As New DataSet Dim result As Byte() conn = New SqlConnection(ConnectionString) insertStatement.Connection = conn If conn.State = ConnectionState.Closed Then conn.Open() End If insertStatement.CommandText = select * from organization  insertStatement.CommandType = CommandType.Text DAdap.SelectCommand = insertStatement DAdap.Fill(DS) 'Dim DS As New DataSet DAdap.Fill(DS) result = CType(DS.Tables(0).Rows(0).Item(imageschk) Byte()) Dim ms As MemoryStream = New MemoryStream(result 0 result.Length) Dim im As Image = Image.FromStream(ms) Picture.Image = im Its helpful to you

Lalit M · Answer

show here good articles step by step guide ...