creating datagrid with code

[David C]

I am trying to dynamically create a datagrid with code that looks exactly
like what the asp tags show below. It has only one column and that column
is a template column with a link button control.

This is how far I was able to take this.

DataGrid dgRecords = new DataGrid();
dgRecords.DataKeyField = "RecordID";
dgRecords.AutoGenerateColumns = false;

TemplateColumn temcol = new TemplateColumn();

I know it's not very much. I just can't figure out how to have the embedded
link button control in the column. Thanks.

<asp:DataGrid id="dgRecords" runat="server" DataKeyField="RecordID"
AutoGenerateColumns="False"
Visible="False">
<Columns>
<asp:TemplateColumn>
<ItemTemplate>
<asp:LinkButton ID="lnkLink" Runat="server">
<%#DataBinder.Eval(Container.DataItem, "Entry")%>
</asp:LinkButton>
</ItemTemplate>
</asp:TemplateColumn>
</Columns>
</asp:DataGrid>

[Teemu Keiski]

Hi,

you would need to implement a class describing the template. See:

Implementing Dynamic ItemTemplates
http://www.tripleasp.net/tutorial.aspx?NavID=27

Second way it to declare template(s) in user controls and use
Page.LoadTemplate to load it from UC.

--
Teemu Keiski
ASP.NET MVP, AspInsider
Finland, EU

"David C" <pr********@yahoo.com> wrote in message
news:Xb*******************@twister.socal.rr.com...

I am trying to dynamically create a datagrid with code that looks exactly
like what the asp tags show below. It has only one column and that column
is a template column with a link button control.

This is how far I was able to take this.

DataGrid dgRecords = new DataGrid();
dgRecords.DataKeyField = "RecordID";
dgRecords.AutoGenerateColumns = false;

TemplateColumn temcol = new TemplateColumn();

I know it's not very much. I just can't figure out how to have the
embedded
link button control in the column. Thanks.

<asp:DataGrid id="dgRecords" runat="server" DataKeyField="RecordID"
AutoGenerateColumns="False"
Visible="False">
<Columns>
<asp:TemplateColumn>
<ItemTemplate>
<asp:LinkButton ID="lnkLink" Runat="server">
<%#DataBinder.Eval(Container.DataItem, "Entry")%>
</asp:LinkButton>
</ItemTemplate>
</asp:TemplateColumn>
</Columns>
</asp:DataGrid>