Monday, August 15, 2011

Cross Page posting in ASP.NET 2.0

Posted by Rahul Kharde at 10:06 PM
System.Web.UI.WebControls.IButtonControl interface contains a new property called PostBackUrl which points to the page to which the current page will postback, Button, ImageButton and LinkButton implements this interface and exposes the cross page postback functionality.

When user clicks the button the current page will postback to the specified page which can access the source page controls through Page.PreviousPage property which returns the reference of previous page, once got the reference of previous page you can use the FindControl method to get the reference of particular or you can expose public properties from source page to provide the type safe access i.e.


Cross Page posting or cross page postback is used to submit a form on one page (say FirstPage.aspx) and retrieve values of controls of this page on another page (say ResultPage.aspx)

There are two ways we can use cross page
postsbacks in ASP.NET


<html xmlns="">
<head id="Head1" runat="server">
    <title>Untitled Pagetitle>
<form id="form1" runat="server">
        <b>User Name : b><asp:TextBox ID="txtUserName" runat="server">asp:TextBox>
        <br />
        <b>Password : b><asp:TextBox ID="txtPassword" runat="server"TextMode="Password">asp:TextBox>
        <br />
        <asp:Button ID="Button1" runat="server" onclick="Button1_Click" PostBackUrl="~/ResultPage.aspx" Text="Submit" />



TextBox txtUserName = (TextBox)Page.PreviousPage.FindControl("txtUserName");
            TextBox txtPassword = (TextBox)Page.PreviousPage.FindControl("txtPassword");
            Response.Write("Welcome:" + txtUserName.Text.ToString());
page posting is yet another new feature introduced in ASP.NET 2.0, which eases the life of developers previously they have to use Server.Transfer which has its own advantages and disadvantages but now this is a part of ASP.NET which results flexibility and efficiency.

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