C# 动态开关箱

我正在尝试为几个不同的用户创建一个简单的switch case控制台菜单：

admin

、

主持人

、以及

用户

<“代码”>“管理”将具有

创建、删除、修改、显示

功能、

版主

-

创建、修改、显示

函数和

user

-

create，show

函数可从中选择

管理开关盒：

if(userType == "admin")
{
    string i = Console.ReadLine();
    switch(i):
    case "create": Console.WriteLine("Created");
                   break;
    case "modify": Console.WriteLine("Modified");
                   break;
    case "delete":Console.WriteLine("Deleted");
                  break;
    case "show":Console.WriteLine("Showed");
                break;
    default: Console.WriteLine("Default");
             break;
}

if(userType == "moderator")
{
    string i = Console.ReadLine();
    switch(i):
    case "create": Console.WriteLine("Created");
                   break;
    case "modify": Console.WriteLine("Modified");
                   break;
    case "show": Console.WriteLine("Showed");
                 break;
    default: Console.WriteLine("Default");
             break;
}

if(userType == "user")
{
    string i = Console.ReadLine();
    switch(i):
    case "create": Console.WriteLine("Created");
                   break;
    case "show": Console.WriteLine("Showed");
                 break;
    default: Console.WriteLine("Default");
             break;
}

慢化剂开关箱：

if(userType == "admin")
{
    string i = Console.ReadLine();
    switch(i):
    case "create": Console.WriteLine("Created");
                   break;
    case "modify": Console.WriteLine("Modified");
                   break;
    case "delete":Console.WriteLine("Deleted");
                  break;
    case "show":Console.WriteLine("Showed");
                break;
    default: Console.WriteLine("Default");
             break;
}

if(userType == "moderator")
{
    string i = Console.ReadLine();
    switch(i):
    case "create": Console.WriteLine("Created");
                   break;
    case "modify": Console.WriteLine("Modified");
                   break;
    case "show": Console.WriteLine("Showed");
                 break;
    default: Console.WriteLine("Default");
             break;
}

if(userType == "user")
{
    string i = Console.ReadLine();
    switch(i):
    case "create": Console.WriteLine("Created");
                   break;
    case "show": Console.WriteLine("Showed");
                 break;
    default: Console.WriteLine("Default");
             break;
}

用户交换机情况：

if(userType == "admin")
{
    string i = Console.ReadLine();
    switch(i):
    case "create": Console.WriteLine("Created");
                   break;
    case "modify": Console.WriteLine("Modified");
                   break;
    case "delete":Console.WriteLine("Deleted");
                  break;
    case "show":Console.WriteLine("Showed");
                break;
    default: Console.WriteLine("Default");
             break;
}

if(userType == "moderator")
{
    string i = Console.ReadLine();
    switch(i):
    case "create": Console.WriteLine("Created");
                   break;
    case "modify": Console.WriteLine("Modified");
                   break;
    case "show": Console.WriteLine("Showed");
                 break;
    default: Console.WriteLine("Default");
             break;
}

if(userType == "user")
{
    string i = Console.ReadLine();
    switch(i):
    case "create": Console.WriteLine("Created");
                   break;
    case "show": Console.WriteLine("Showed");
                 break;
    default: Console.WriteLine("Default");
             break;
}

---

有没有办法把这些开关盒做成一个动态开关？如果我在思考或解释错误，请纠正我。

这是一个很好的候选人

在策略模式中，功能由可以传递的接口对象表示。不同的实现允许动态更改行为

例如：

interface ConsoleInteractor
{
    void performAction(string action);
}

class UserConsoleInteractor : ConsoleInteractor
{
    public void performAction(string action)
    {
        switch(i)
        {
            case "create": 
                Console.WriteLine("Created");
                break;
            case "show":
                Console.WriteLine("Showed");
                break;
            default: 
                Console.WriteLine("Default");
                break;
        }
    }
}

切换（无双关语）在每个情况下检查角色。如果不允许这样做，跳过它

string i = Console.ReadLine();
if (allowed(userType, i)){
  switch(i):
  case "create": Console.WriteLine("Created");
      handleCreate();
  break;
  case "show":Console.WriteLine("Showed");
      handleShow();
  break;
  case "delete":Console.WriteLine("Deleted");
      handleDelete();
  break;
  default: Console.WriteLine("Default");
    handleDefault(userType);
  break;
}

---

你最好有一张功能或动作的地图

var actionsAdmin = new Dictionary<string, Action>{
  {"create", ()=>Console.WriteLine("create")}
  {"modify", ()=>Console.WriteLine("modify")}
}

var actionsUser = new Dictionary<string, Action>{
  {"show", ()=>Console.WriteLine("show")}
  {"foodle", ()=>Console.WriteLine("foodle")}
}

---

开关案例的动态等价物是字典查找。例如：

Dictionary<string, Action> userActions = {
        { "create", () => Console.WriteLine("created") },
        { "show", () => Console.WriteLine("showed") } };
Dictionary<string, Action> adminActions = {
        { "create", () => Console.WriteLine("created") },
        { "show", () => Console.WriteLine("showed") },
        { "delete", () => Console.WriteLine("deleted") } };

Dictionary<string, Dictionary<string, Action>> roleCapabilities = {
        { "user", userActions },
        { "administrator", adminActions } };

roleCapabilities[userType][action]();

---

英雄所见略同；）+1尽管我认为另一本查这个角色的字典（“用你的话选择正确的地图”）是有用的。所有的代码重复都会带来麻烦。将访问检查置于交换机外部，并传入变量

i

，而不是字符串文本。否则，有一天在执行大规模擦除时，您会意外地为读取权限编写测试代码。您是对的，我的重点是集中在单个点上进行访问控制。在这个过程中，引入了我所担心的错误类型。现已修复：）