reactjs - 如何在 Typescript 中为 React useReducer Hook 操作创建类型定义？

Question

假设我们有 userReducer定义如下:

function userReducer(state: string, action: UserAction): string {
  switch (action.type) {
    case "LOGIN":
      return action.username;
    case "LOGOUT":
      return "";
    default:
      throw new Error("Unknown 'user' action");
  }
}

定义UserAction 的最佳方式是什么？输入以便可以调用 dispatch两者都有 username有效载荷并且没有:

dispatch({ type: "LOGIN", username: "Joe"}});
/* ... */
dispatch({ type: "LOGOUT" });

如果类型是这样定义的:

type UserActionWithPayload = {
  type: string;
  username: string;
};

type UserActionWithoutPayload = {
  type: string;
};

export type UserAction = UserActionWithPayload | UserActionWithoutPayload;

在 "LOGIN"情况下，编译器在 reducer 中抛出错误:TS2339: Property 'username' does not exist on type 'UserAction'.   Property 'username' does not exist on type 'UserActionWithoutPayload'.
如果使用可选成员定义类型:

export type UserAction = {
  type: string;
  username?: string;
}

然后编译器显示此错误:TS2322: Type 'string | undefined' is not assignable to type 'string'.   Type 'undefined' is not assignable to type 'string'.
这里缺少什么？也许整个方法是错误的？

项目使用 TypeScript 3.8.3 和 React.js 16.13.0。

Answer 1

经过数小时的挖掘和试验，通过 Typescript enum 找到了一个非常优雅的解决方案和联合类型的行动:

enum UserActionType {
  LOGIN = "LOGIN",
  LOGOUT = "LOGOUT"
}

type UserState = string;

type UserAction =
| { type: UserActionType.LOGIN; username: string }
| { type: UserActionType.LOGOUT }

function userReducer(state: UserState, action: UserAction): string {
  switch (action.type) {
    case UserActionType.LOGIN:
      return action.username;
    case UserActionType.LOGOUT:
      return "";
    default:
      throw new Error();
  }
}

function App() {
  const [user, userDispatch] = useReducer(userReducer, "");

  function handleLogin() {
    userDispatch({ type: UserActionType.LOGIN, username: "Joe" });
  }

  function handleLogout() {
    userDispatch({ type: UserActionType.LOGOUT });
  }

  /* ... */
}

使用上述方法没有错误或警告，此外还有一个非常严格的操作使用契约(Contract)。