Introduction:
In ASP.NET 1.1 a page can only PostBack to itself, it is a common technique to post the page to another page. In ASP.NET 2.0 ASPX page can post to different ASPX page by using the button's PostBackUrl property. Where ASPX is the extension of ASP.NET web application.
We can see the button's property in the following figure.
Figure 1: Button's PostBackUrl property.
Steps for implementing the cross page PostBack using controls in ASP.NET 2.0:
Step 1: Open the new website in Microsoft Visual studio 2005.
Step 2: Drop a button control on the form and you can also drag other controls according the need of application.
Step 3: Set the PostBackUrl property of the button where you want to PostBack.
For Example: Let us suppose we want to PostBack to another page from "My Page" then we have to pass the PostBackUrl as follows:
PostBackUrl="http://www.c-sharpcorner.com".
Now you can see the source code and also set the PostBackUrl of that page where you want to post the page.
MyPage.aspx:
<%@ Page Language="C#" AutoEventWireup="true" CodeFile="Default.aspx.cs" 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>My Page</title>
</head>
<body>
<form id="form1" runat="server">
<div>
<table cellpadding="0" cellspacing="0" border="2" height="50px" style="width: 33%">
<tr>
<td align="center" valign="top" style="padding-top:20px;">
<asp:TextBox ID="TxtName" runat="server"></asp:TextBox><br/><br/>
<asp:Button ID="BtnSubmit" runat="server" PostBackUrl="http://www.c-sharpcorner.com" Text="Submit" OnClick="Button1_Click" />
</td>
</tr>
</table>
</div>
</form>
</body>
</html>
Design view of the above source code:
Figure 2: My first page.
When you post to another page in ASP.NET 2.0, you can reach out to the values in the page using the following methods:
MyPage.aspx.cs:
using System;
using System.Data;
using System.Configuration;
using System.Web;
using System.Web.Security;
using System.Web.UI;
using System.Web.UI.WebControls;
using System.Web.UI.WebControls.WebParts;
using System.Web.UI.HtmlControls;
public partial class _Default : System.Web.UI.Page
{
protected void Page_Load(object sender, EventArgs e)
{}
protected void Button1_Click(object sender, EventArgs e)
{
if (PreviousPage != null)
{
TextBox txtName = (TextBox)
PreviousPage.FindControl("txtName");
this.TxtName.Text= txtName.Text;
}
}
}
Output: After debug the .cs file you will get the output as:
Figure 3: My Page is display.
After that click on Submit button "My Page" will PostBack to another page. Here we have set the PostBackUrl="http://www.c-sharpcorner.com", so the following page will open that mean the page will post to c-sharpcorner.com.
Figure 4: New page is open.
PreviousPage property holds the posted page. We check if it is null and use the FindControl() method to reach our control.