[RESOLVED] [2005]Setting a datagrid index

[space_monkey]

Here's what i'm currently doing. To bind a dataset to a datagrid.

Dim myReports As New DATA_ACCESS
            Dim dsReports As New DataSet
            dsReports = myReports.GET_MY_REPORTS(Session("USER_NAME"))
            ReportGrid.DataSource = dsReports
            ReportGrid.DataBind()
            myReports = Nothing
            dsReports = Nothing

There's no problem there. Everything comes back fine. My question is, "how do i set the index/key for the grid?". Something like what i did with a drop down box.

Dim dsDepts As New DataSet
            dsDepts = depts.GET_DEPT_OPTS
            ddDept.DataSource = dsDepts
            ddDept.DataTextField = "DEPT_INFO_NAME"
            ddDept.DataValueField = "DEPT_INFO_IDNM"
            ddDept.DataBind()

Here when an item is selected in the dropdown box i can see a behind the scenes value for it. I'd like to do the same thing with the datagrid. Basically have a column in my dataset that doesn't show in the datagrid but is recognized as the index for that row in the code behind. Or am i going about this the wrong way.

Thanks
Andy (ASP.NET noob)

[HoraShadow]

You can setup an entire column of Ids and make it visible="False" in the HTML part of the Datagrid, using templates. Only the code behind sees it.

HTH
HoraShadow

[space_monkey]

Thanks for your reply. I was on this path already for a hypertext column i wanted; i just didn't realize that it extended to this.

here's what i ended up doing. I first had to turn off the AutoGeneratecolumns and then just provide a template for each of the columns that i wanted after that. For example

<asp:TemplateField>   
                                    <HeaderTemplate>
                                        Author
                                    </HeaderTemplate> 
                                    <ItemTemplate>
                                            <%#(Container.DataItem("Author"))%>
                                    </ItemTemplate>
                            </asp:TemplateField>

It seems to be working quite nicely.

Thanks