我有以下ListView：
<asp:ListView ID="procedureTicketList" runat="server" ...
<ItemTemplate>
<asp:GridView ID="MyGridView" runa...
如何以编程方式访问MyGridView？
答案 0 :(得分：2)
使用FindControl方法。
答案 1 :(得分：1)
尝试以下代码段。
protected void procedureTicketList_DataBound(object sender, EventArgs e)
{
GridView gv= ((GridView)e.Item.FindControl("MyGridView"));
.
.
.
}
修改
检查以下代码段。
<%@ Page Language="C#" AutoEventWireup="true" CodeFile="Default2.aspx.cs" Inherits="Default2" %>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head runat="server">
<title></title>
</head>
<body>
<form id="form1" runat="server">
<div>
<asp:ListView runat="server" ID="_simpleTableListView" OnItemDataBound="_simpleTableListView_ItemDataBound">
<LayoutTemplate>
<table>
<thead>
<tr>
<th>
ID
</th>
<th>
Title
</th>
</tr>
</thead>
<tbody>
<asp:PlaceHolder runat="server" ID="itemPlaceholder" />
</tbody>
</table>
</LayoutTemplate>
<ItemTemplate>
<tr>
<td>
<%# Eval("ID") %>
</td>
<td>
<%# Eval("title") %>
</td>
<td>
<asp:GridView ID="MyGridView" runat="server">
</asp:GridView>
</td>
</tr>
</ItemTemplate>
</asp:ListView>
</div>
</form>
</body>
</html>
using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.UI;
using System.Web.UI.WebControls;
public partial class Default2 : System.Web.UI.Page
{
protected void Page_Load(object sender, EventArgs e)
{
_simpleTableListView.DataSource = new Movie().GetAll;
_simpleTableListView.DataBind();
}
protected void _simpleTableListView_ItemDataBound(object sender, ListViewItemEventArgs e)
{
GridView gv= ((GridView)e.Item.FindControl("MyGridView"));
}
}
public class Movie
{
public int Id { get; set; }
public string Title { get; set; }
public List<Movie> GetAll
{
get
{
return new List<Movie>()
{
new Movie{Id=1,Title="A"},
new Movie{Id=2,Title="B"},
};
}
}
}