RxJS: BehaviorSubject and groupBy operator

Question

Here it says :

// RxJS v6+
import { from } from 'rxjs';
import { groupBy, mergeMap, toArray } from 'rxjs/operators';

const people = [
  { name: 'Sue', age: 25 },
  { name: 'Joe', age: 30 },
  { name: 'Frank', age: 25 },
  { name: 'Sarah', age: 35 }
];
//emit each person
const source = from(people);
//group by age
const example = source.pipe(
  groupBy(person => person.age),
  // return each item in group as array
  mergeMap(group => group.pipe(toArray()))
);
/*
  output:
  [{age: 25, name: "Sue"},{age: 25, name: "Frank"}]
  [{age: 30, name: "Joe"}]
  [{age: 35, name: "Sarah"}]
*/
const subscribe = example.subscribe(val => console.log(val));

In my code I don't create an observable with the 'from' operator, but rather with BehaviorSubject.asObservable() method.

Person { name: string, age: number }

    private _all: BehaviorSubject<Person[]>;
    all: Observable<Person[]>;

    constructor() {
        this._all = new BehaviorSubject<Person[]>([]);
        this.all = this._all.asObservable();
    }

I can iterate over 'all' with the async pipe. But when I try to use the groupBy operator I get the array itself, instead of the containing persons one by one as a stream :

this.all.pipe(
    groupBy(
        item => ...   <-- here 'item' is Person[], not a Person
    )
);

What am I doing wrong ?

score 2 · Accepted Answer · answered Aug 01 '18 at 04:59

The simple answer is (unfortunately) that it is not possible this way. Here is another post concerning a similar question.

You could do a step in between to reach the wanted goal:

Option 1: without subscription

// directly access the BehaviorSubject's value
const list = from(this._all.value);
this.list.pipe(
    groupBy(
        item => ...  
    )
);

Option 2: with subscription, as it is an Observable

// catch the plain list inside the subscription
this.all.subscribe(result => {
    const list = from(result);
    this.list.pipe(
        groupBy(
            item => ... 
        )
    );
});

RxJS: BehaviorSubject and groupBy operator

1 Answers1