typescript - 在 TypeScript 中切换特定类型

Question

我有一个接口(interface)Action:

interface Action {}

以及此Action SpecificAction 的实现:

class SpecificAction implements Action {
   payload?: Any
}

是否可以在TS中构造一个switch操作符，像这样:

let action: Action
switch (action) {
   case SpecificAction: //it works
       console.log(action.payload) // it doesn't 
}

在那种情况下是否有可能知道该操作已经是 SpecificAction 类型？

Answer 1

目前看来有一些选择，但都有一些缺点

• 受歧视的工会 docs stackblitz ，但你需要一个专用属性作为鉴别器

interface Action {}

class SpecificAction implements Action {
  kind: "specific";
  payload?: any;
}

class ToggleAction implements Action {
  kind: "toggle";
  toggle: boolean;
}

let action: SpecificAction | ToggleAction;
switch (action.kind) {
  case "specific":
    console.log(action.payload) // it works 
    break;
  case "toggle":
    console.log(action.toggle) // it works 
    break;        
}

• 用户定义的类型保护 docs stackblitz , 但你需要 if 语句而不是 switch

interface Action {}

class SpecificAction implements Action {
  payload?: any;
}

class ToggleAction implements Action {
  toggle: boolean;
}

let isSpecific = (p: any): p is SpecificAction => !!p.payload
let isToggle = (p: any): p is ToggleAction => !!p.toggle

let action: Action;
if (isSpecific(action)) {
  console.log(action.payload) // it works 
} else if (isToggle(action)) {
  console.log(action.toggle) // it works 
}

• 构造函数属性 github stackblitz , 但您暂时需要转换为所需的类型

interface Action { }

class SpecificAction implements Action {
  payload?: any;
}

class ToggleAction implements Action {
  toggle: boolean;
}

switch (action.constructor) {
  case SpecificAction:
    console.log((<SpecificAction>action).payload) // it kinda works 
    break;
  case ToggleAction:
    console.log((<ToggleAction>action).toggle) // it kinda works 
    break;
  }