如何获得对状态树的访问? (@ ngrx/效果2.x) [英] How to get access of the state tree in effects? (@ngrx/effects 2.x)

问题描述

我正在将 @ ngrx/effects 从1.x更新到2.x

I am updating @ngrx/effects from 1.x to 2.x

在1.x版本中，我实际上可以访问状态树:

In 1.x I have access of state tree in effect:

  constructor(private updates$: StateUpdates<AppState>) {}

  @Effect() bar$ = this.updates$
    .whenAction(Actions.FOO)
    .map(obj => obj.state.user.isCool)
    .distinctUntilChanged()
    .filter(x => x)
    .map(() => ({ type: Actions.BAR }));

现在在2.x中，它只给我动作.还有没有办法访问状态树?还是我应该避免这样使用，因为这不是一个好习惯?

Now in 2.x, it only gives me action. Is there still a way to get access of the state tree? Or should I avoid using like this because it is not a good practice?

  constructor(private actions$: Actions) {}

  @Effect() bar$ = this.actions$
    .ofType(ActionTypes.FOO)
    .map((obj: any) => {
      console.log(obj);              // here is action only
      return obj.state.user.isCool   // so it is wrong here
    })
    .distinctUntilChanged()
    .filter(x => x)
    .map(() => ({ type: ActionTypes.BAR }));

推荐答案

另一种方法是使用.withLatestFrom(this.store).因此，完整的代码是:

Another way is using .withLatestFrom(this.store). So the complete code is:

  constructor(
    private actions$: Actions,
    private store: Store<AppState>
  ) {}

 @Effect() bar$ = this.actions$
    .ofType(ActionTypes.FOO)
    .withLatestFrom(this.store, (action, state) => state.user.isCool)
    .distinctUntilChanged()
    .filter(x => x)
    .map(() => ({ type: ActionTypes.BAR }));

这篇关于如何获得对状态树的访问? (@ ngrx/效果2.x)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！