Hi,

      How to check the object is null in C#. For the code below

TextBox tbox = new TextBox();

 You can use the "as" keyword to try to cast the object. If the casting fails null is returned.

tbox =Page.PreviousPage.FindControl("textboxFirstName") as TextBox;

like this:

TextBox tbox = (TextBox)Page.PreviousPage.FindControl("textboxFirstName");
if (tbox == null) {
    // it was null
}

 Go for "as" as suggest by azam.

Use

if(tbox!=null)

{

//do your code if control finds.

}

else

{

//tbox is null/

}

tgopi99:

tbox = (TextBox)Page.PreviousPage.FindControl("textboxFirstName");

if( tbox!=null)

{

}

best rEgard

pAthan 

Hi,

   I am getting the following error when i am checking null value of an object as you said. Here is my code

Object reference not set to an instance of an object.

Description: An unhandled exception occurred during the execution of the current web request. Please review the stack trace for more information about the error and where it originated in the code.

Exception Details: System.NullReferenceException: Object reference not set to an instance of an object.

Source Error:

Line 39:                 TextBox tbox = new TextBox();
Line 40:                 DropDownList dlist = new DropDownList();
Line 41:                 tbox = (TextBox)Page.PreviousPage.FindControl("textboxFirstName") as TextBox;
Line 42:                 if (tbox != null)
Line 43:                 {

I even checked with the above code, but still it is giving the following error.

Object reference not set to an instance of an object.

Description: An unhandled exception occurred during the execution of the current web request. Please review the stack trace for more information about the error and where it originated in the code.

Exception Details: System.NullReferenceException: Object reference not set to an instance of an object.

Source Error:

Line 39:                 TextBox tbox = new TextBox();
Line 40:                 DropDownList dlist = new DropDownList();
Line 41:                 tbox = (TextBox)Page.PreviousPage.FindControl("textboxFirstName") as TextBox;
Line 42:                 if (tbox != null)
Line 43:                 {

 First check that the PreviousPage property is not null and also make sure that the Previous Page consists of the element named textboxFirstName. You can also expose properties in the PreviousPage to access the controls instead of using FindControl method.

 Here is a small example that will clear up the things further:

This is the code for my previousPage:

  <asp:TextBox ID="txtName" runat="server" />
        <asp:Button ID="Button1" runat="server" PostBackUrl="~/FirstPage.aspx" Text="Button" OnClick="Button1_Click" />

And here is the page that access the PreviousPage.

 protected void Page_Load(object sender, EventArgs e)
        {
            if (PreviousPage != null)
                {
                    TextBox tb = PreviousPage.FindControl("txtName") as TextBox;
                    Response.Write(tb.Text);
                }
           
            if (!Page.IsPostBack)
            {
                           
            }
        }

If you are not able to find the control in the PreviousPage then you can use properties in the Previous to access the value of the control. Like the following:

public class MyPreviousPage : ....

public string Name { return txtName.Text; }

Now, in the main page you can access like this:

MyPreviousPage.Name

You will also need to add the following directive in the main page (NOT Previous Page)

<%@ PreviousPageType VirtualPath="~/MyPreviousPage.aspx"  %>
 

Very useful one, Thank you very much.