如何在MVC 4 C＃网站中创建动态菜单和子菜单 [英] HOW can Creating Dynamic Menus and Submenus in an MVC 4 C# Website

问题描述

如何在MVC 4 C＃网站中创建动态菜单和子菜单正确的示例和编码



我想要动态创建的菜单结构

 <   ul  >  
 
 <   li  >  @ Html.ActionLink（主页，索引，主页）<   / li  >  
 <   li  >  @ Html.ActionLink（Home，Index，Home）<   / li  >  
 <   li  >  @ Html.ActionLink（ASP.NET，ASPPage，ASPNET）
 <   ul  >  
 <   li  >  @ Html.ActionLink（ASP.NET，ASPPage， ASPNET）<   / li  >  
 <   li  > @Html.ActionLink(\"ASP.NET\",\"ASPPage\",\"ASPNET\")<   / li  >  
 <   li  >  @ Html.ActionLink（ASP.NET，ASPPage，ASPNET）<   / li  >  
 <   li  >  @ Html.ActionLink（ASP。 NET，ASPPage，ASPNET）<   / li  >  
 <   / ul  >  
 <   / li  >  
 <   li  >  @ Html.ActionLink（主页，索引，主页）<   / li  >  
 <  <跨度 class =code-leadattribute> li  >  @ Html.ActionLink（Home，Index，Home）<   / li  >  
 <   li  >  @ Html.ActionLink（ 主页，索引，主页）<   / li  >  
 <   li  >  @ Html.ActionLink（Home，Index，Home）<   / li  >  
 
 <   / ul  >  

解决方案

我已经在下面解释了动态创建示例菜单结构。

创建一个类并添加一个属性来存储li结构。如果你需要UI集合也声明这个。

  public   class 菜单
 {
 < span class =code-keyword> public  List< string> li { get ;  set ;} 
 
} 

创建一个ActionResult

  public  ActionResult Index（）
 {
  var  menu =  new 菜单（）; 
 menu.li = Loadli（）; 
  return  view（）; 
} 

  public 列表< string> LoadUl（）
 {
  var  list = List< string>（）; 
 list.Add（  Home）; 
 list.Add（  ASP.NET）; 
} 

创建一个返回Menu的视图。如果您正在使用任何其他类，则在该类上声明li属性，根据你的结构循环浏览项目。

 @ model菜单
 {
 
< ; UL> 
 @foreach（ var  item  in  Model.li）
 {
< li> @item < /   li  > ;  
} 
 < /   ul  >  
} 

希望这会有所帮助

创建一个具有以下字段的模型名称菜单

  public   int  MenuId { get ;  set ;} 
  public   int  ParentMenuId { get ;  set ;} 
  public   str ing  MenuName { get ;  set ;} 
  public   string 查看{获取;  set ;} 
  public   string 控制器{ get ;  set ;} 
  public   int 订单{获取; 设置;} 

 @ model菜单
 {
 
< ul> 
 @foreach（ var  item  in  Model）
 {
< ; li> @ Html.ActionLink（  @ Model.View，  @ Model.Controller，  @ Model.MenuName< / li> 
} 
< / ul> 
}  

HOW can Creating Dynamic Menus and Submenus in an MVC 4 C# Website proper example and coding

Menu Structure that i want to create dynamically

<ul>
       
    <li>  @Html.ActionLink("Home","Index","Home")</li>
    <li> @Html.ActionLink("Home","Index","Home")</li>
    <li>@Html.ActionLink("ASP.NET","ASPPage","ASPNET")
        <ul>
        <li>@Html.ActionLink("ASP.NET","ASPPage","ASPNET")</li>
        <li>@Html.ActionLink("ASP.NET","ASPPage","ASPNET")</li>
        <li>@Html.ActionLink("ASP.NET","ASPPage","ASPNET")</li>
        <li>@Html.ActionLink("ASP.NET","ASPPage","ASPNET")</li>
        </ul>
    </li>
    <li>  @Html.ActionLink("Home","Index","Home")</li>
    <li> @Html.ActionLink("Home","Index","Home")</li>
    <li>  @Html.ActionLink("Home","Index","Home")</li>
    <li> @Html.ActionLink("Home","Index","Home")</li>
   
    </ul>

解决方案

I have explained below for creating a sample menu structure dynamically.
Create a class and add a property for storing li structure.If u need UI collection declare this too.

public class Menu
{
public List<string> li{get;set;}

}

Create an ActionResult

public ActionResult Index()
{
    var menu=new Menu();
menu.li=Loadli();
return view();
}

public List<string> LoadUl()
{
var list=List<string>();
list.Add("Home");
list.Add("ASP.NET");
}

Create a view which return Menu.If you are using anyother class then declare the li property on that class and loop through the item according to your structure.

@model Menu
{

<ul>
  @foreach (var item in Model.li)
            {
                <li>@item</li>
            }
</ul>
}

Hope this helps

Create One Model Name Menu having following fields

public int MenuId{get;set;}
public int ParentMenuId{get;set;}
public string MenuName{get;set;}
public string View{get;set;}
public string Controller{get;set;}
public int Order{get;set;}

@model Menu
{
 
<ul>
  @foreach (var item in Model)
            {
                <li>@Html.ActionLink("@Model.View","@Model.Controller","@Model.MenuName</li>
            }
</ul>
}

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