Question

我想知道表达类似消息（值对象）的惯用方法是什么我们假设我要发送一些消息：UserEnter，UserLeft，UserSendGreeting等＆＃39;。我希望以下内容可以表达我的困境

我认为有两个选项可以表达它：

trait UserAction
case class UserEnter (username:String,hour:Long,day:WeekDay,room:Int)
case class UserLeft (username:String,hour:Long,day:WeekDay,room:Int)
case class UserSendGreeting (username:String,,hour:Long,day:WeekDay,greeting:String)

def foo(userActions:List[UserAction]) = {
userActions match {
case UserEnter(userName)::tail =>
sendWelcomeMessage(userName)
handleOtherActions(tail) 

case UserLeft::tail => "must enter first"
case _ => "do something else"
}

另一种方式是：

    trait Action
    case class Enter(room:Int) extends Action
    case object Left(room:Int) extends Action
    case object Greet(msg:String) extends Action

    case class UserAction(username:String,action:Action,,hour:Long,day:WeekDay)

    def foo(userActions:List[UserAction]) = {
    userActions match {
    case head::tail if head.action == Enter =>
    sendWelcomeMessage(head.userName)    
    handleOtherActions(tail)             
    case head::tail if head.action == Left => "must enter first"
    case _ => "do something else"
    }

第一个选项更明确，更容易进行模式匹配，但更重复的代码。另一个对动作本身更明确，但模式匹配更详细哪种方式更好？

Answer 1

不要过分复杂化。如果您需要一对User（在您的案例中为字符串名称）和Action，请使用Tuple。收集所有＆＃34; ...其他属性＆＃34;也是有意义的。在相应的Action内。

type User = String

sealed trait Action

case class Enter(room:Int) extends Action

case class Left(room:Int) extends Action

case class Greet(msg:String) extends Action


def foo(userActions:List[(User, Action)]) = userActions match {
    case (username, _:Enter)::tail =>
      sendWelcomeMessage(username)
      handleOtherActions(tail)
    case (username, _:Left)::tail => "must enter first"
    case _ => "do something else"
  }

另外请不要忘记，模式匹配允许您unapply任意深度的层次结构：

case (username, Enter(room))::tail =>

Answer 2

我认为1种方法更好，但您可以对UserAction特性进行约束以获得用户名字段：

trait UserAction {
  def username: String
}
case class UserEnter(username:String, room:Int)
case class UserLeft(username:String, room:Int)
case class UserSendGreeting(username:String, greeting:String)

这种方式很容易

1）模式匹配

2）使用通用UserAction获取用户名字段

Answer 3

如何：TABLES: VBRK. DATA: BEGIN OF it_test, BUKRS(5), "longer version of VBRK-BUKRS, FKDAT LIKE VBRK-FKDAT, END OF it_test. DATA: tt_test TYPE STANDARD TABLE OF it_test. * I would strongly recommend to set a filter! SELECT * FROM VBRK INTO CORRESPONDING FIELD OF it_test. IF it_test-BUKRS = 'xxxx'. it_test-BUKRS = 'XXXXX'. APPEND it_test to tt_test. ENDIF. ENDSELECT.

case class UserAction(user: User, action: Action)

甚至更好

def foo(actions: Seq[UserActions]) = actions match {
    case UserAction(user, _: EnterAction)::tail => 
      processEnter(user)
      handleOtherActions(tail)
    case UserExit(user, _: ExitAction)::tail => processExit(user)
    ...
}

或者，也许，只是

def foo(actions: Seq[UserAction]) = actions.find { ua => 
  !ua.action.process(ua.user)
}

表达类似值对象的惯用scala / FP方式

3 个答案: