[RxJS] Combining Streams with CombineLatest

Posted 2020-06-20 Answer1215

Two streams often need to work together to produce the values you’ll need. This lesson shows how to use an input stream and an interval stream together and push an object with both values through the stream.

 

const Observable = Rx.Observable;

const startButton = document.querySelector(‘#start‘);
const halfButton = document.querySelector(‘#half‘);
const quarterButton = document.querySelector(‘#quarter‘);
const input = document.querySelector("#input");
const stopButton = document.querySelector(‘#stop‘);
const resetButton = document.querySelector(‘#reset‘);


const data = {count:0};
const inc = (acc)=> ({count: acc.count + 1});
const reset = (acc)=> data;


const start$ = Observable.fromEvent(startButton, ‘click‘);
const half$ = Observable.fromEvent(halfButton, ‘click‘);
const quarter$ = Observable.fromEvent(quarterButton, ‘click‘);
const stop$ = Observable.fromEvent(stopButton, ‘click‘);
const reset$ = Observable.fromEvent(resetButton, ‘click‘);
const starters$ =  Observable.merge(
    start$.mapTo(1000),
    half$.mapTo(500),
    quarter$.mapTo(250)
 );
const intervalActions = (time) => {
      return Observable.merge(
        Observable.interval(time)
        .takeUntil(stop$)
        .mapTo(inc),
        reset$.mapTo(reset)
      )
    };
const timer$ = starters$
 .switchMap(intervalActions)
 .startWith(data)
 .scan( (acc, curr) => curr(acc));
const input$ = Observable.fromEvent(input, "input")
  .map(ev=>ev.target.value);

Observable.combineLatest(
  timer$,
  input$
)
// We will get an array combineLatest
.map((array)=>{
  return {count: array[0].count, input: array[1]}
})
.subscribe(x => console.log(x))

 

They last param of combineLatest is a result function, which can parse the result in the fucntion:

Observable.combineLatest(
  timer$,
  input$,
  (timer, input)=>{
     return {count: timer.count, input}
  }
)
.subscribe(x => console.log(x))