Hi all viewers...

 i have visited many articles but they all converge to selecting a gridview and display its content on another grid view but on the same page.

Any method i can use to pass it to another page?

i tried

 <asp:TemplateField><HeaderTemplate>SMS Update</HeaderTemplate>
               <ItemTemplate>
               <asp:HyperLink runat="server" Text="Get" 
NavigateUrl="~/Update.aspx?CompanyName=<%# container.dataitem ("CName")%>"></asp:HyperLink>
               </ItemTemplate>
               </asp:TemplateField>    

 But it gives me error in the gridview stating that "the server tag is  not well formed"

 
Any idea or solution to this scenario?
 

refer here

http://www.koffeekoder.com/ArticleDetails.aspx?id=133_Passing_GridView_Values 

Thanks for your reply Vinz. 

The site was fruitful. I can pass values to the next page

Musafiir:

<asp:TemplateField><HeaderTemplate>SMS Update</HeaderTemplate> <ItemTemplate> <asp:HyperLink runat="server" Text="Get" NavigateUrl="~/Update.aspx?CompanyName=<%# container.dataitem ("CName")%>"></asp:HyperLink> </ItemTemplate> </asp:TemplateField>

 <asp:TemplateField><HeaderTemplate>SMS Update</HeaderTemplate> <ItemTemplate>
               <asp:HyperLink runat="server" Text="Get"
NavigateUrl='~/Update.aspx?CompanyName=<%# container.dataitem ("CName")%>'></asp:HyperLink>
               </ItemTemplate>
               </asp:TemplateField>

<asp:TemplateField><HeaderTemplate>SMS Update</HeaderTemplate>
               <ItemTemplate>
               <asp:HyperLink runat="server" Text="Get" 
NavigateUrl='<%# "~/Update.aspx?CompanyName=" + Server.UrlEncode(DataBinder.Eval(Container.DataItem,"CName").ToString())%>'></asp:HyperLink>               </ItemTemplate>
               </asp:TemplateField>

Thanks for all your contributions :-)

 Musafiir

Software Developer