combineLatest deprecated in favor of static combineLatest

Question

After running the rxjs migration tool using

rxjs-5-to-6-migrate -p src/tsconfig.app.json

I'm now getting a linting error:

combineLatest is deprecated: Deprecated in favor of static combineLatest.

Here is my code before running the migration command:

this.store.combineLatest(         this.store.select(lang.getCurrent),         this.store.select(lang.getCurrentLocale)     ).subscribe(([state, currentLang, locale]) => {         this._language = session.language === currentLang ? '' : currentLang;         this._locale = session.locale === locale ? '' : locale;     }); 

My code after running the migration command: (currently presenting a linting error)

import {map, combineLatest} from 'rxjs/operators'; this.store.combineLatest(         this.store.select(lang.getCurrent),         this.store.select(lang.getCurrentLocale)     ).subscribe(([state, currentLang, locale]) => {         this._language = session.language === currentLang ? '' : currentLang;         this._locale = session.locale === locale ? '' : locale;     }); 

The question was asked in this stackoverflow questions, but it was not specific enough: Angular 6 ng lint duplicate errors and warnings, combineLatest is deprecated .

Answer 1

In rxjs 6.5+

import { combineLatest } from 'rxjs';  combineLatest([a$, b$, c$]);  

And for most applications it's helpful to map the array of observables to a new value as well:

combineLatest([a$, b$, c$]).pipe(   map(([a$, b$, c$]) => ({     a: a$,      b: b$,      c: c$   })) ); 

Also see: https://www.learnrxjs.io/learn-rxjs/operators/combination/combinelatest

Answer 2

Deprecated!

Please refer to ofir fridman's answer for the correct syntaxs as of RxJs 6.5

---

I found an answer in this article titled: RxJS 6: What's new and what has changed? ( which comes from official docs):

The solution is to convert:

import { combineLatest } from 'rxjs/operators';  a$.pipe(combineLatest(b$, c$)); 

into:

import { combineLatest } from 'rxjs';  combineLatest([a$, b$, c$]);