Create Template column dynamically in DataGrid

Dolly Jain

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7 Mar 20061 min read

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This articles helps in creating template columns dynamically with bound columns using a template class

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Introduction

This Article helps in creation of template columns dynamically . I got a condition when i have to display few column values from database and few template (checkbox) column dynamically.

I have created a sample template class which is required in project.Firstly modify template class as per your requirement.As here i have given checkbox as template column example.

Steps to implement code in your project :

1) Set the DataGrid AutoGenerateColumns property to false .

2) Example to Bound the static Columns

<asp:DataGrid id="ItemsGrid" runat="server" OnItemCommand="Grid_CartCommand" AutoGenerateColumns="False"
CellPadding="3" BorderWidth="1px" BorderColor="Black">
<HeaderStyle BackColor="#C0C0FF"></HeaderStyle>
<Columns>
<asp:ButtonColumn Text="Add" ButtonType="PushButton" HeaderText="Add to cart" CommandName="AddToCart"></asp:ButtonColumn>
<asp:ButtonColumn Text="Remove" ButtonType="PushButton" HeaderText="Remove from cart" CommandName="RemoveFromCart"></asp:ButtonColumn>
<asp:BoundColumn DataField="StringValue" HeaderText="Item"></asp:BoundColumn>
<asp:BoundColumn DataField="CurrencyValue" HeaderText="Price" DataFormatString="{0:c}">
<ItemStyle HorizontalAlign="Right"></ItemStyle>
</asp:BoundColumn>
</Columns>
</asp:DataGrid>

3) In Page_load add following Code

if (!IsPostBack)
{
// Load this data only once.
ItemsGrid.DataSource= CreateDataSource(); //Function to create dynamic column
ItemsGrid.DataBind();
}

4) In OnInit() add following code protected void OnInit(EventArgs e)// // CODEGEN: This call is required by the ASP.NET Web Form Designer. //

CreateDataGridColumn();

InitializeComponent();

}

base.OnInit(e);

//-----------------------------------------------------------------------------

public

{

override

{

void CreateDataGridColumn()// Create dynamic column to add to Columns collection. //-----------------------------------------------------------------------------

TemplateColumn tc1 =

//-----Class used named DataGridTempla.cs------

tc1.HeaderTemplate =

tc1.ItemTemplate =

tc1.EditItemTemplate =

tc1.FooterTemplate =

ItemsGrid.Columns.Add(tc1);

new TemplateColumn();new DataGridTempla(ListItemType.Header, "Select1");new DataGridTempla(ListItemType.Item, "Select1");new DataGridTempla(ListItemType.EditItem, "");new DataGridTempla(ListItemType.Footer, "");//-----------------------------------------------------------------------------

TemplateColumn tc2=

tc2.HeaderTemplate=

tc2.ItemTemplate=

ItemsGrid.Columns.Add(tc2);

new TemplateColumn();new DataGridTempla(ListItemType.Header,"Select2");new DataGridTempla(ListItemType.Item,"Select2");//-----------------------------------------------------------------------------

BoundColumn NumberColumn =

NumberColumn.HeaderText="Item Number";

NumberColumn.DataField="IntegerValue";

new BoundColumn();// Add column to Columns collection.

ItemsGrid.Columns.AddAt(2, NumberColumn);

}

//-----------------------------------------------------------------------------

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Written By

Dolly Jain

Web Developer

India

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' note i'am using for test only access database db1.mdb. ' it have just one table whit a,b,c columns and some random data, ' just for teste. Imports System.Data.OleDb Imports System.Data Public Class WebForm1 Inherits System.Web.UI.Page #Region " Web Form Designer Generated Code " 'This call is required by the Web Form Designer. <System.Diagnostics.DebuggerStepThrough()> Private Sub InitializeComponent() End Sub Protected WithEvents myDataGrid As System.Web.UI.WebControls.DataGrid Protected strcon As String = "Jet OLEDB:Database Password=;Data Source=C:\Inetpub\wwwroot\testAzamGrid\db1.mdb;Password=;Provider=Microsoft.Jet.OLEDB.4.0;User ID=Admin;" Protected dt As DataTable = New DataTable 'NOTE: The following placeholder declaration is required by the Web Form Designer. 'Do not delete or move it. Private designerPlaceholderDeclaration As System.Object Private Sub Page_Init(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Init 'CODEGEN: This method call is required by the Web Form Designer 'Do not modify it using the code editor. InitializeComponent() End Sub #End Region Private Sub Page_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load BuildDataGrid() End Sub Private Sub BuildDataGrid() ' Fill dt GetData("a asc") For i As Integer = 0 To dt.Columns.Count - 1 ' Creating Template Column Dim test As TemplateColumn = New TemplateColumn Dim columnName As String = dt.Columns(i).ColumnName test.HeaderTemplate = New DataGridTemplate(ListItemType.Header, columnName) 'Adding the Rows to the DataGrid For j As Integer = 0 To dt.Rows.Count - 1 Dim score As String = dt.Rows(j)(i).ToString() test.ItemTemplate = New DataGridTemplate(ListItemType.Item, score) Next myDataGrid.Columns.Add(test) Next myDataGrid.DataSource = dt myDataGrid.DataBind() End Sub Private Function GetData(ByRef campo As String) As DataTable Dim cnn As OleDbConnection = New OleDbConnection(strcon) cnn.Open() Dim cmd As OleDbCommand = cnn.CreateCommand() cmd.CommandType = System.Data.CommandType.Text cmd.CommandText = "select a,b,c from TB_NOME order by " & campo Dim ad As OleDbDataAdapter = New OleDbDataAdapter(cmd) 'Dim dt As DataTable = New DataTable ' local in use. ad.Fill(dt) cnn.Close() Return dt End Function End Class Public Class DataGridTemplate Inherits System.Web.UI.Page Implements ITemplate Public templateType As ListItemType Public columnName As String Public Sub New(ByVal type As ListItemType, ByVal colname As String) templateType = type columnName = colname End Sub Public Sub InstantiateIn(ByVal container As System.Web.UI.Control) Implements System.Web.UI.ITemplate.InstantiateIn Dim lc As Literal = New Literal Dim lb As LinkButton = New LinkButton Select Case templateType Case ListItemType.Header lc.Text = "" + columnName + "" lb.Text = "Edit" lb.CommandName = "EditButton" container.Controls.Add(lb) container.Controls.Add(lc) Exit Select Case ListItemType.Item lc.Text = "Item " + columnName container.Controls.Add(lc) Exit Select Case ListItemType.EditItem Dim tb As TextBox = New TextBox tb.Text = "" container.Controls.Add(tb) Exit Select Case ListItemType.Footer lc.Text = "" + columnName + "" container.Controls.Add(lc) Exit Select End Select End Sub End Class //////////////////////////////// WebForm1.aspx <%@ Page Language="vb" AutoEventWireup="false" Codebehind="WebForm1.aspx.vb" Inherits="testAzamGrid.WebForm1"%> <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>WebForm1</title> <meta name="GENERATOR" content="Microsoft Visual Studio .NET 7.1"> <meta name="CODE_LANGUAGE" content="Visual Basic .NET 7.1"> <meta name=vs_defaultClientScript content="JavaScript"> <meta name=vs_targetSchema content="http://schemas.microsoft.com/intellisense/ie5"> </head> <body MS_POSITIONING="GridLayout"> <form id="Form1" method="post" runat="server"> <asp:DataGrid ID="myDataGrid" GridLines="Both" AutoGenerateColumns="false" Visible="True" ShowHeader="True" Runat="server"></asp:DataGrid> </form> </body> </html> /////////////////////////////////////////////////

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Create Template column dynamically in DataGrid

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