Question

我是RxJS的新手。我想创建一个AppState对象的可观察对象，可以随时更改，并订阅它以获得这些更改。这是一个精简的实现：

export class AppState {

  public get observable(): Observable<any> {
     return Observable.of(this._state);
  }
}


// appState is injected into my component via angular DI

this.appState.observable
     .subscribe((appState) => {
        console.log('appState: ', appState);)
     }, (err) => {
        console.log('Error: ' + err);
     }, () =>{
        console.log('Completed');
     });

但它只运行一次并立即调用completed。因此，当我更改appState时，订阅已经结束。

如何让订阅永远保持活力，例如KnockOutJS风格。这用于Angular应用程序

更新：我部分使用Subject。但问题是现在它发出了许多相同价值的副本。

// full appState.ts
import { Injectable } from '@angular/core';
import { Observable, Subject, BehaviorSubject } from 'rxjs';

export type InternalStateType = {
   [key: string]: any
};

@Injectable()
export class AppState {

   public _state: InternalStateType = {};
   public subject: Subject<any>;

   constructor() {
      this.subject = new Subject();
   }

   /**
    * Return an observable for subscribing to.
    */
   public get observable() {
      return this.subject;
   }

   /**
    * Return a clone of the current state.
    */
   public get state() {
      this._state = this._clone(this._state);
      this.subject.next(this._state);
      return this._state;
   }

   /**
    * Never allow mutation
    */
   public set state(value) {
      throw new Error('do not mutate the `.state` directly');
   }

   public get(prop?: any) {
      /**
       * Use our state getter for the clone.
       */
      const state = this.state;
      return state.hasOwnProperty(prop) ? state[prop] : state;
   }

   public set(prop: string, value: any) {
      /**
       * Internally mutate our state.
       */
      return this._state[prop] = value;
   }

   private _clone(object: InternalStateType) {
      /**
       * Simple object clone.
       */
      return JSON.parse(JSON.stringify(object));
   }
}

需要更改哪些内容才能让每次更改只发出一次更改this._state？

Answer 1

您需要使用Subject或BehaviorSubject。

主题将在订阅者传递给Subject时向其发出值，而BehaviorSubject将在订阅时发出最后一个值，然后在值可用时继续发出值。

从保持活动的对象创建observable

1 个答案: