asp.net 多级数据绑定菜单

发布于 2024-12-19 02:00:27 字数 786 浏览 2 评论 0原文

我目前正在使用 asp.net 菜单控件从表父/子项加载。我遇到的问题是，如果这个孩子还有另一个孩子。从这个意义上说，我的代码是静态的，我似乎找不到更好的或“该”的方法来做到这一点。我已将站点地图视为数据源，但我不需要站点地图，并且觉得这对于我需要实现的目标来说太过分了。

foreach (ClassName option in list)
{
   MenuItem module = new MenuItem(option.Description.ToLower(), "", "", option.Url + "?option=" + option.Optionid);
   module.Selectable = true;
   navigation.Items.Add(module);
   //this is my second level
   foreach (ClassName child in listfromparent(option.Optionid))
   {
         MenuItem childmenu = new MenuItem(child.Description.ToLower(), "", "", child.Url + "?option=" + child.Optionid);
         module.ChildItems.Add(childmenu);
   }
 }

正如你所看到的，这有效，但只适用于 2 个级别:( 当然，我可以在 child 中放入另一个子级别来创建第三级别，但是如果有第四、第五级别怎么办？所以这就是为什么我需要它自己做。我注意到树视图有 onpopulate 但显然菜单没有。提前致谢。

原文

I am currently using an asp.net menu control to load from a table parent/child items. The problem I am having is that if the child has another child. My code is kindof static in that sense and I can't seem to find a better or "the" way to do it. I have seen sitemap as datasources but i don't need a sitemap and feel that would just be overkill for what I need to achieve.

foreach (ClassName option in list)
{
   MenuItem module = new MenuItem(option.Description.ToLower(), "", "", option.Url + "?option=" + option.Optionid);
   module.Selectable = true;
   navigation.Items.Add(module);
   //this is my second level
   foreach (ClassName child in listfromparent(option.Optionid))
   {
         MenuItem childmenu = new MenuItem(child.Description.ToLower(), "", "", child.Url + "?option=" + child.Optionid);
         module.ChildItems.Add(childmenu);
   }
 }

as you can see this works but for 2 levels :(
and of course i could put another childlevel inside child to create the 3rd level but what if there is a 4th, 5th? So that's why I need it to do it itself. I noticed treeview has onpopulate but apparently Menu doesn't. Thanks in advance.

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肤浅与狂妄 2024-12-26 02:00:27

这是您可以做到的一种方法。

使用邻接列表表示表中的父/子关系
将邻接列表映射到树结构
将该树结构转换为菜单项结构

也许您可以跳过中间步骤并将邻接列表直接映射到 MenuItems 树在 MenuItem 上有一些扩展方法。

但无论如何...

Default.aspx

<%@ Page Language="C#" Inherits="MenuTreeDemo.Default" %>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html>
<head runat="server">
    <title>Default</title>
</head>
<body>
    <form id="form1" runat="server">
        <asp:Menu ID="MyMenu" runat="server" StaticDisplayLevels="3" />
    </form>
</body>
</html>

Default.aspx.cs

using System;
using System.Web;
using System.Web.UI;
using System.Web.UI.WebControls;
using System.Data;
using System.Collections.Generic;

namespace MenuTreeDemo
{
    public partial class Default : System.Web.UI.Page
    {
        protected void Page_Load(object sender, EventArgs e)    
        {
            if (!IsPostBack)
            {
                MenuNode root = ConvertTableToTree(GetTreeTable());
                foreach (MenuNode topLevelNode in root.Children)
                {
                    MyMenu.Items.Add(topLevelNode.ToMenuItem()); // Visits all nodes in the tree.
                }
            }
        }


        // The menu tree as an adjacency list in a table. 
        static DataTable GetTreeTable()
        {
            DataTable table = new DataTable();
            table.Columns.Add("Id", typeof(int));
            table.Columns.Add("Description", typeof(string));
            table.Columns.Add("Url", typeof(string));
            table.Columns.Add("ParentId", typeof(int));

            table.Rows.Add(1, "TopMenu1", "/foo.html", 0);
            table.Rows.Add(2, "SubMenu1.1", "/baz.html", 1);
            table.Rows.Add(3, "SubMenu1.2", "/barry.html", 1);
            table.Rows.Add(4, "SubMenu1.2.1", "/skeet.html", 3);
            table.Rows.Add(5, "TopMenu2", "/bar.html", 0);
            table.Rows.Add(6, "TopMenu3", "/bar.html", 0);
            table.Rows.Add(7, "SubMenu3.1", "/ack.html", 6);

            return table;
        }


        // See e.g. http://stackoverflow.com/questions/2654627/most-efficient-way-of-creating-tree-from-adjacency-list
        // Assuming table is ordered.
        static MenuNode ConvertTableToTree(DataTable table)
        {
            var map = new Dictionary<int, MenuNode>();
            map[0] = new MenuNode() { Id = 0 }; // root node

            foreach (DataRow row in table.Rows)
            {
                int nodeId = int.Parse(row["Id"].ToString());
                int parentId = int.Parse(row["ParentId"].ToString());

                MenuNode newNode = MenuNodeFromDataRow(row);

                map[parentId].Children.Add(newNode);
                map[nodeId] = newNode;
            }

            return map[0]; // root node
        }


        static MenuNode MenuNodeFromDataRow(DataRow row)
        {
            int nodeId = int.Parse(row["Id"].ToString());
            int parentId = int.Parse(row["ParentId"].ToString());
            string description = row["Description"].ToString();
            string url = row["Url"].ToString();

            return new MenuNode() { Id=nodeId, ParentId=parentId, Description=description, Url=url };
        }
    }
}

MenuNode.cs

using System;
using System.Collections.Generic;
using System.Web.UI.WebControls;

namespace MenuTreeDemo
{
    public class MenuNode
    {
        public int Id { get; set; }
        public int ParentId { get; set; }
        public string Description { get; set; }
        public string Url { get; set; }
        public List<MenuNode> Children { get; set; }


        public MenuNode ()
        {
            Children = new List<MenuNode>();
        }


        // Will visit all descendants and turn them into menu items.
        public MenuItem ToMenuItem()
        {
            MenuItem item = new MenuItem(Description) { NavigateUrl=Url };
            foreach (MenuNode child in Children)
            {
                item.ChildItems.Add(child.ToMenuItem());
            }

            return item;
        }
    }
}

Here's one way you could do it.

Represent parent/child relationship in your table with an adjacency list
Map that adjacency list into a tree structure
Convert that tree structure into your structure of menu items

Maybe you could skip that middle step and map the adjacency list straight to a tree of MenuItems, maybe with some extension methods on MenuItem.

But anyway...

Default.aspx

<%@ Page Language="C#" Inherits="MenuTreeDemo.Default" %>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html>
<head runat="server">
    <title>Default</title>
</head>
<body>
    <form id="form1" runat="server">
        <asp:Menu ID="MyMenu" runat="server" StaticDisplayLevels="3" />
    </form>
</body>
</html>

Default.aspx.cs

using System;
using System.Web;
using System.Web.UI;
using System.Web.UI.WebControls;
using System.Data;
using System.Collections.Generic;

namespace MenuTreeDemo
{
    public partial class Default : System.Web.UI.Page
    {
        protected void Page_Load(object sender, EventArgs e)    
        {
            if (!IsPostBack)
            {
                MenuNode root = ConvertTableToTree(GetTreeTable());
                foreach (MenuNode topLevelNode in root.Children)
                {
                    MyMenu.Items.Add(topLevelNode.ToMenuItem()); // Visits all nodes in the tree.
                }
            }
        }


        // The menu tree as an adjacency list in a table. 
        static DataTable GetTreeTable()
        {
            DataTable table = new DataTable();
            table.Columns.Add("Id", typeof(int));
            table.Columns.Add("Description", typeof(string));
            table.Columns.Add("Url", typeof(string));
            table.Columns.Add("ParentId", typeof(int));

            table.Rows.Add(1, "TopMenu1", "/foo.html", 0);
            table.Rows.Add(2, "SubMenu1.1", "/baz.html", 1);
            table.Rows.Add(3, "SubMenu1.2", "/barry.html", 1);
            table.Rows.Add(4, "SubMenu1.2.1", "/skeet.html", 3);
            table.Rows.Add(5, "TopMenu2", "/bar.html", 0);
            table.Rows.Add(6, "TopMenu3", "/bar.html", 0);
            table.Rows.Add(7, "SubMenu3.1", "/ack.html", 6);

            return table;
        }


        // See e.g. http://stackoverflow.com/questions/2654627/most-efficient-way-of-creating-tree-from-adjacency-list
        // Assuming table is ordered.
        static MenuNode ConvertTableToTree(DataTable table)
        {
            var map = new Dictionary<int, MenuNode>();
            map[0] = new MenuNode() { Id = 0 }; // root node

            foreach (DataRow row in table.Rows)
            {
                int nodeId = int.Parse(row["Id"].ToString());
                int parentId = int.Parse(row["ParentId"].ToString());

                MenuNode newNode = MenuNodeFromDataRow(row);

                map[parentId].Children.Add(newNode);
                map[nodeId] = newNode;
            }

            return map[0]; // root node
        }


        static MenuNode MenuNodeFromDataRow(DataRow row)
        {
            int nodeId = int.Parse(row["Id"].ToString());
            int parentId = int.Parse(row["ParentId"].ToString());
            string description = row["Description"].ToString();
            string url = row["Url"].ToString();

            return new MenuNode() { Id=nodeId, ParentId=parentId, Description=description, Url=url };
        }
    }
}

MenuNode.cs

using System;
using System.Collections.Generic;
using System.Web.UI.WebControls;

namespace MenuTreeDemo
{
    public class MenuNode
    {
        public int Id { get; set; }
        public int ParentId { get; set; }
        public string Description { get; set; }
        public string Url { get; set; }
        public List<MenuNode> Children { get; set; }


        public MenuNode ()
        {
            Children = new List<MenuNode>();
        }


        // Will visit all descendants and turn them into menu items.
        public MenuItem ToMenuItem()
        {
            MenuItem item = new MenuItem(Description) { NavigateUrl=Url };
            foreach (MenuNode child in Children)
            {
                item.ChildItems.Add(child.ToMenuItem());
            }

            return item;
        }
    }
}

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