The ASP.NET version of the Server.Transfer method accepts a second argument which, if true, causes the query string and all the form's fields to be transferred and made available to the destination page. However, when the destination page executes, the following error is likely to be generated:
The View State is invalid for this page and might be corrupted
The reason for this error is that the EnableViewStateMac attribute of the
element of Web.config is set to true by default. This causes a message authentication check (MAC) on the view state of the page being posted back. The view state is held in the __VIEWSTATE hidden field, and the MAC fails because the MAC only checks each page. In other words, the view state of the page where Server.Transfer is called isn't valid on the destination page. This problem is discussed more estensively on the KB article mentioned at the bottom of this page.
You avoid this problem by not passing True as the second argument of the Server.Transfer method. However, you must devise alternative ways to make the query string and controls' values of the source page available to the destination page, if these values must be processed by the destination page. There are two ways to achive this. The easiest way is to make all controls on the source page public, as in:
Public Class WebForm1
Public WithEvents txtUserName As System.Web.UI.WebControls.TextBox
' ...
End Class
A safer way is to wrap these control's values in properties:
Public Class WebForm1
Private WithEvents txtUserName As System.Web.UI.WebControls.TextBox
' ...
ReadOnly Property UserName() As String
Get
Return Me.txtUserName.Text
End Get
End Property
End Class
In both cases you can then access the actual value from the destination page as follows:
' inside the destination page
Dim sourcePage As WebPage1 = DirectCast(Context.Handler, WebForm1)
Dim userName As String = sourcePage.UserName
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