Asp.net Web Forms GridView动态列发布

Question

我们有一个4.7.2版本的.Net网页表单应用程序。当我们以编程方式将列插入到GridView组件中时，TemplateFields在回发后不会呈现。我们找到了一个关于这个问题的主题。我认为这是一个.Net框架错误。帖子上有一个反馈网址，但我打不开那个页面。

FeedBack url：

https://connect.microsoft.com/VisualStudio/feedback/ViewFeedback.aspx?FeedbackID=104994&wa=wsignin1.0

主题url：

https://forums.asp.net/t/1102255.aspx?GridView+Columns+Insert+Problems

示例代码：

Aspx ->

 <asp:Content ID="BodyContent" ContentPlaceHolderID="MainContent" runat="server">
   <asp:GridView ID="TestGrid" runat="server" AutoGenerateColumns="false">
       <Columns>
           <asp:BoundField HeaderText="Col1" DataField="Name" />
           <asp:TemplateField>
               <HeaderTemplate>
                   <asp:Label ID="Col2HeaderLabel" runat="server" Text="Col2"></asp:Label>
               </HeaderTemplate>
               <ItemTemplate>
                   <asp:CheckBox ID="Col2CheckBox" runat="server" />
               </ItemTemplate>
           </asp:TemplateField>
           <asp:BoundField HeaderText="Col3" DataField="Surname" />
       </Columns>
   </asp:GridView>

    <asp:Button ID="TestButton" runat="server" Text="Test" />
</asp:Content>

Aspx.cs ->

        protected override void OnInit(EventArgs e)
        {
            base.OnInit(e);

            TestGrid.Columns.Insert(0,new BoundField { HeaderText = "Dynamic Col", DataField = "Description" });
        }

        protected void Page_Load(object sender, EventArgs e)
        {
            TestGrid.DataSource = GetList();
            TestGrid.DataBind();
        }

        private List<Info> GetList()
        {
            List<Info> list = new List<Info>();
            list.Add(new Info { Description = "Row 1", Name = "Name 1", Surname = "Surname 1" });
            list.Add(new Info { Description = "Row 2", Name = "Name 2", Surname = "Surname 2" });
            list.Add(new Info { Description = "Row 3", Name = "Name 3", Surname = "Surname 3" });

            return list;
        }
    }

    public class Info
    {
        public string Name { get; set; }
        public string Surname { get; set; }
        public string Description { get; set; }
    }

Answer 1

@yusuf，您需要将数据源重新绑定到PostBack上的GridView。将数据gridView.DataSource保存到会话变量中，然后在PostBack上再次执行绑定。这将确保绑定不会丢失。

protected void Page_Load(object sender, EventArgs e)
{
if (Session["gvDS"] != null && IsPostBack)
    {
        gridView1.DataSource = Session["gvDS"];
        gridView1.DataBind();
    }
    else
        BindGridView();


}

private void BindGridView()
{
  // Your SQL statements go here, etc, then:

  gridView1.DataSource = YourDataSetTable;
  gridView1.DataBind();

  Session["gvDS"] = gridView1.DataSource;  // save into Session

}