typescript - typescript 区分联合的 switch 语句的替代方法

Question

我已创建 this playground这是代码:

type BundlerError = Error;
type BundlerWarning = Error;

export type BundlerState =
  | { type: 'UNBUNDLED' }
  | { type: 'BUILDING'; warnings: BundlerWarning[] }
  | { type: 'GREEN'; path: string;  warnings: BundlerWarning[] }
  | { type: 'ERRORED'; error: BundlerError }

const logEvent = (event: BundlerState) => {
    switch (event.type) {
      case 'UNBUNDLED': {
        console.log('received bundler start');
        break;
      }
      case 'BUILDING':
        console.log('build started');
        break;
      case 'GREEN':
        if(event.warnings.length > 0) {
          console.log('received the following bundler warning');

          for (let warning of event.warnings) {
              warning
            console.log(warning.message);
          }
        }
        console.log("build successful!");
        console.log('manifest ready');
        break;
      case 'ERRORED':
        console.log("received build error:");
        console.log(event.error.message);
        break;
    }
}

BundlerState 是一个有区别的联合，并且 switch 缩小了类型。
问题是它不能扩展，而且大扩展的 switch 语句非常可怕。
有没有更好的方法可以写这个并且仍然保持好的类型缩小？
你不可以做这个:

const eventHandlers = {
  BUNDLED: (event: BundlerState) => event.type // type is not narrowed
  // etc,
};

const logEvent = (event: BundlerState) => eventHandlers['BUNDLED'](event);

因为类型没有缩小。

Answer 1

这是我经常使用的一种模式(或其变体)。

type BundlerStatesDef = {
   UNBUNDLED: {}
   BUILDING: { warnings: BundlerWarning[] }
   GREEN: { path: string; warnings: BundlerWarning[] }
   ERRORED: { error: BundlerError }
}
type BundlerStateT = keyof BundlerStatesDef
type BundlerStates = { [K in BundlerStateT]: { type: K } & BundlerStatesDef[K] }
type BundlerHandler<K extends BundlerStateT> = (params: BundlerStates[K]) => void
type BundlerHandlers = { [K in BundlerStateT]: BundlerHandler<K> }

使用上面定义的类型，您可以有一个非常符合人体工程学的实现，如下所示:

const handlers: BundlerHandlers = {
  UNBUNDLED: params => console.log(params),
  BUILDING: params => console.log(params),
  GREEN: params => console.log(params),
  ERRORED: params => console.log(params)
}

const logEvent = <E extends BundlerStateT>(event: BundlerStates[E]) =>
  (handlers[event.type] as BundlerHandler<E>)(event)

PLAYGROUND

贴近你的原始定义，甚至不那么冗长，你可以这样做:

type BundlerError = Error
type BundlerWarning = Error

export type BundlerState =
  | { type: 'UNBUNDLED' }
  | { type: 'BUILDING'; warnings: BundlerWarning[] }
  | { type: 'GREEN'; path: string;  warnings: BundlerWarning[] }
  | { type: 'ERRORED'; error: BundlerError }

export type BundlerHandlers = { [K in BundlerState['type']]: (params: Extract<BundlerState, { type: K }>) => void }

const handlers: BundlerHandlers = {
  UNBUNDLED: params => console.log(params),
  BUILDING: params => console.log(params),
  GREEN: params => console.log(params),
  ERRORED: params => console.log(params)
}

const logEvent = (event: BundlerState) =>
  (handlers[event.type] as (params: Extract<BundlerState, { type: typeof event['type'] }>) => void )(event)

PLAYGROUND