SIGN UP MEMBER LOGIN:    
Blog

Random display of records in gridview

Posted by Satyapriya Nayak Blogs | ASP.NET Controls in C# Sep 12, 2011
In this article we will know how to get random values from the database, bind with the Gridview when user clicks refresh link every time.

Table structure

table structure in sql

Default.aspx code

<%@ Page Language="VB" AutoEventWireup="false" CodeFile="Default.aspx.vb" Inherits="_Default" %> 
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
 
<
html xmlns="http://www.w3.org/1999/xhtml">
<head runat="server">
    <title>Untitled Page</title>   
     <link href="style.css" rel="stylesheet" type="text/css" />
 
</
head>
<
body>
    <form id="form1" runat="server">
    <div>
   
    </div>
    <asp:GridView ID="GridView1" runat="server">
        <HeaderStyle BackColor="#FFCC66" />
        <AlternatingRowStyle BackColor="#FF6666" />
    </asp:GridView>
    <table>
    <tr>
    <td><a href="Default.aspx" id="uu">Refresh to get random values</a></td>
    </tr>
    </table>
    </form>
</body>
</
html>

Default.aspx.vb code

Imports System.Data.SqlClient
Imports System.Data
Partial Class _Default
    Inherits System.Web.UI.Page
    Dim strConnString As String = System.Configuration.ConfigurationManager.ConnectionStrings.Item("ConnectionString").ToString()
    Dim con As New SqlConnection(strConnString)
    Dim sqlda As SqlDataAdapter
    Dim com As SqlCommand
    Dim ds As DataSet
    Dim str As String
    Protected Sub Page_Load(ByVal sender As Object, ByVal e As System.EventArgs) Handles Me.Load
        If Not IsPostBack Then
            bindgrid()
        End If
    End Sub
    Sub bindgrid()
        Try
            con.Open()
            str = "select * from student"
 
           com = New SqlCommand(str, con)
            sqlda = New SqlDataAdapter(com)
            ds = New DataSet
            sqlda.Fill(ds, "student")
            Dim dt As DataTable = ds.Tables(0)
            dt.Columns.Add(New DataColumn("RandomNumber", GetType(Integer)))
            Dim i As Integer
            Dim r1 As New Random()
            For i = 0 To dt.Rows.Count - 1
                dt.Rows(i)("RandomNumber") = r1.Next(10000)
            Next i
            Dim dv As DataView = dt.DefaultView
            dv.Sort = "RandomNumber"
            GridView1.DataSource = dv
            GridView1.DataBind()
            con.Close()
        Catch ex As Exception
            Response.Write(ex.Message)
        End Try
    End Sub
End Class

Output

grid view in asp.net

share this blog :
post comment