Status Twitter Blog: How to find control from footer row of the GridView in ASP.NET

Rabu, 04 September 2013

How to find control from footer row of the GridView in ASP.NET

If you want to add Control to the footer Row in GridView, you must use footer template in GridView also set ShowFooter property is true. Also if you find control in code you can use FindControl menthod . Use single line for accessing control in code

lets take an example

Suppose your footer row take one TextBox control with id property (id="nametxt")

</FooterTemplate>

Now find TextBox control in code with simple example

var nametextbox =GridView1.FooterRow.FindControl("nametxt")as TextBox;

Now again your footer row take one Label control with label ID property (ID="namelbl")

</FooterTemplate>

Now find Label control in code with simple example

var lbl =GridView1.FooterRow.FindControl("namelbl")as Label;

Example of How to insert item from GridView footer Row
Table Structure

How to find control from footer row of the GridView in ASP.NET

Database table

<%@ Page Language="C#" EnableEventValidation ="false" AutoEventWireup="true" CodeFile="additem.aspx.cs" Inherits="additem" %>

<!DOCTYPE html>

<html xmlns="http://www.w3.org/1999/xhtml">
<head runat="server">
<title></title>
</head>
<body>
<form id="form1" runat="server">
<div>
<asp:GridView ID="GridView1" runat="server" ShowFooter ="true" AutoGenerateColumns ="false" OnRowCommand="GridView1_RowCommand">
<Columns >

<asp:TemplateField HeaderText ="Press button">
<ItemTemplate >

<asp:Label ID="Label1" runat="server" Text='<%# Eval("namet") %>'></asp:Label>

</ItemTemplate>
<FooterTemplate >
<asp:Button ID="bt" runat ="server" Text ="Insert item" CommandName ="InsertRecord"/>

</FooterTemplate>
</asp:TemplateField>
<asp:TemplateField HeaderText ="Insert value">
<FooterTemplate >
<asp:TextBox ID="txtCustomerID" runat="server" Text="" />

</FooterTemplate>

</asp:TemplateField>

</Columns>

</asp:GridView>
<asp:Label ID="result" runat="server" Text=""></asp:Label>
</div>
</form>
</body>
</html>

Codebehind

using System;
using System.Collections.Generic;
using System.Configuration;
using System.Linq;
using System.Web;
using System.Web.UI;
using System.Web.UI.WebControls;

public partial class additem : System.Web.UI.Page
{
protected void Page_Load(object sender, EventArgs e)
{

if (!Page .IsPostBack)
{
binddata();
}
}
public void binddata()
{
System.Data.SqlClient.SqlConnection con = new System.Data.SqlClient.SqlConnection();
con.ConnectionString = ConfigurationManager.ConnectionStrings["ConnectionString"].ToString();
con.Open();
System.Data.SqlClient.SqlCommand cmd = new System.Data.SqlClient.SqlCommand();
cmd.CommandText = "Select * from [footerex]";
cmd.Connection = con;
System.Data.SqlClient.SqlDataAdapter da = new System.Data.SqlClient.SqlDataAdapter(cmd);
System.Data.DataSet ds = new System.Data.DataSet();
da.Fill(ds);
GridView1.DataSource = ds;
GridView1.DataBind();
}

protected void GridView1_RowCommand(object sender, GridViewCommandEventArgs e)
{
if (e.CommandName=="InsertRecord")
{
var nametx = GridView1.FooterRow.FindControl("namebox") as TextBox;
System.Data.SqlClient.SqlConnection con = new System.Data.SqlClient.SqlConnection(@"Data Source=(LocalDB)\v11.0;AttachDbFilename=|DataDirectory|\Database.mdf;Integrated Security=True");
con.Open();
System.Data.SqlClient.SqlCommand cmd = new System.Data.SqlClient.SqlCommand();
cmd.Parameters.AddWithValue("@name", nametx .Text);
cmd.CommandText = "insert into [footerex](namet)values(@name)";
cmd.Connection = con;
int a = cmd.ExecuteNonQuery();
if (a > 0)
{
result.Text = "Row Updated";
con.Close();
binddata();

}
else
{
result.Text = "failed";
}
}

}
}

Output
Before insertion

After Insertion

Status Twitter Blog

Rabu, 04 September 2013

How to find control from footer row of the GridView in ASP.NET

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