RxJS and optional inner query - rxjs

I have a kind of web form which has 2 parts (master and optional child) and want to save the form values
A first call is for master data. The call result is used as a parameter for second call. But in some cases we don't need to save child part.
In the end of saving in the subscribe section I want to show "Success" message with one field from the first call result. I mean we can have one or two calls but show the success message when both of them completed
Master Data Call -> Child Data Call (optional) -> Success message
How to implement the behaviour using RxJS in a proper way?
Addition:
Based on comments I created an example here:
https://rxjs.rxplayground.com/
var master = from([1,2])
var child = of(10)
master.pipe(concatMap(m=>(m==1)?child:EMPTY)).subscribe(res=>console.log(res))
(shows 10 but should be 1,2)
Try to change "master" value. In any case log output should match the master. Also it would be nice to avoid "of(m)" inside of concatMap call
One more addition:
I tried this but not sure that it is the best solution
var master = from([1,2]);
var child = of("*");
master.pipe(
mergeMap(m => (m === 1 ? child.pipe(mapTo(m)) : of(m)))
).subscribe(console.log);
(shows 1,2)

I was saying this,just check once
import { from, of } from 'rxjs';
import { concatMap } from 'rxjs/operators';
const master$ = from([1, 2, 3, 4]);
const child$ = of(20);
// const child$ = () => console.log("child called");
master$
.pipe(
concatMap((masterData) => {
if (masterData % 2 === 0) {
child$;
// child$(); // if you want to check,child is being called or not
}
return of(masterData);
})
)
.subscribe(console.log);

This solution does exactly what I need
var master = from([1,2]);
var child = of("*");
master.pipe(
mergeMap(m => (m === 1 ? child.pipe(mapTo(m)) : of(m)))
).subscribe(console.log);
(shows 1,2)

Related

Using RxJS to manipulate a stream of items to obtain an array of streamed items

i'm kinda new to rxjs and can't get my head around this problem:
I have two streams:
one with incoming objects
---a----b----c----d----->
one with the selected object from a list
-------------------c---->
From the incoming objects stream make a stream of the list of objects (with scan operator)
incoming: ----a--------b-------c----------d----------------\>
list: -------[a]----[a,b]----[a,b,c]----[a,b,c,d]---------\>
When a list object is selected (n), start a new stream
the first value of the new stream is the last value of the list sliced ( list.slice(n))
incoming: ----a--------b-------c----------d--------------------e-------->
list: -------[a]----[a,b]----[a,b,c]----[a,b,c,d]--------->
selected object: ---------------------------------c------->
new stream of list: ------[c,d]-----[c,d,e]--->
i can't get the last value of the list stream when the object is selected,,,
made a marble diagram for better understanding,
selectedObject$ = new BehaviorSubject(0);
incomingObjects$ = new Subject();
list$ = incomingObjects$.pipe(
scan((acc, val) => {
acc.push(val);
return acc;
}, [])
)
newList$ = selectedObject$.pipe(
withLastFrom(list$),
switchMap(([index,list])=> incomingObjects$.pipe(
scan((acc, val) => {
acc.push(val);
return acc;
}, list.slice(index))
))
)
A common pattern I use along with the scan operator is passing reducer functions instead of values to scan so that the current value can be used in the update operation. In this case you can link the two observables with a merge operator and map their values to functions that are appropriate - either adding to a list, or slicing the list after a selection.
// these are just timers for demonstration, any observable should be fine.
const incoming$ = timer(1000, 1000).pipe(map(x => String.fromCharCode(x + 65)), take(10));
const selected$ = timer(3000, 3000).pipe(map(x => String.fromCharCode(x * 2 + 66)), take(2));
merge(
incoming$.pipe(map(x => (s) => [...s, x])), // append to list
selected$.pipe(map(x => (s) => { // slice list starting from selection
const index = s.indexOf(x);
return (index !== -1) ? s.slice(index) : s;
}))
).pipe(
scan((list, reducer) => reducer(list), []) // run reducer
).subscribe(x => console.log(x)); // display list state as demonstration.
If I understand the problem right, you could follow the following approach.
The key point is to recognize that the list Observable (i.e. the Observable obtained with the use of scan) should be an hot Observable, i.e. an Observable that notifies independent on whether or not it is subscribed. The reason is that each new stream you want to create should have always the same source Observable as its upstream.
Then, as you already hint, the act of selecting a value should be modeled with a BehaviorSubject.
As soon as the select BehaviorSubject notifies a value selected, the previous stream has to complete and a new one has to be subscribed. This is the job of switchMap.
The rest is to slice the arrays of numbers in the right way.
This is the complete code of this approach
const selectedObject$ = new BehaviorSubject(1);
const incomingObjects$ = interval(1000).pipe(take(10));
const incomingObjectsHot$ = new ReplaySubject<number[]>(1);
incomingObjects$
.pipe(
scan((acc, val) => {
acc.push(val);
return acc;
}, [])
)
.subscribe(incomingObjectsHot$);
selectedObject$
.pipe(
switchMap((selected) =>
incomingObjectsHot$.pipe(
map((nums) => {
const selIndex = nums.indexOf(selected);
if (selIndex > 0) {
return nums.slice(selIndex);
}
})
)
),
filter(v => !!v)
)
.subscribe(console.log);
An example can be seen in this stackblitz.

Filtering a BehaviorSubject

I have a BehaviorSubject that I'd like to be able to filter, but maintain it's behavior-subject-like quality that new subscribers always get a value when they subscribe, even if the last value emitted was filtered out. Is there a succinct way to do that using built-in functions from rxjs? For example:
const isEven = (n) => n % 2 === 0;
const source = new BehaviorSubject(1);
const stream = source.pipe(filter(isEven));
stream.subscribe((n) => console.log(n)); // <- I want this to print `1`
source.next(2); // prints `2`; that's good
source.next(3); // does not print anything; that's good
I've written my own implementation, but would prefer a simpler solution using existing operators instead if it's easy.
Just use a second BehaviorSubject
const { BehaviorSubject } = rxjs;
const { filter} = rxjs.operators;
const isEven = (n) => n % 2 === 0;
const source = new BehaviorSubject(1);
const stream = new BehaviorSubject(source.getValue());
source.pipe(filter(isEven)).subscribe(stream);
stream.subscribe(val => { console.log(val); });
source.next(2);
source.next(3);
<script src="https://cdnjs.cloudflare.com/ajax/libs/rxjs/6.4.0/rxjs.umd.min.js"></script>
Adrian's answer gets the credit, it looks like he answer the best way given the built-in operators available with rxjs itself. It didn't quite meet my needs, so I published my custom operator in my little library s-rxjs-utils. It it called filterBehavior(). From the docs:
Works like filter(), but always lets through the first emission for each new subscriber. This makes it suitable for subscribers that expect the observable to behave like a BehaviorSubject, where the first emission is processed synchronously during the call to subscribe() (such as the async pipe in an Angular template).
Your stream has already been piped to use the isEven filter, so your initial value of 1 is not shown in your console is behaving as expected.
If you want to see your initial value of 1, subscribe directly to the BehaviourSubject:
const isEven = (n) => n % 2 === 0;
const source = new BehaviorSubject(1);
const stream = source.pipe(filter(isEven));
// should print 1, and should print 2 and 3 when your source is nexted.
source.subscribe((n) => console.log(n));
stream.subscribe((n) => console.log(n)); // <- should NOT Print 1, because it has been filtered
source.next(2); // prints `2`; that's good
source.next(3); // does not print anything; that's good

Conditionally producing multiple values based on item value and merging it into the original stream

I have a scenario where I need to make a request to an endpoint, and then based on the return I need to either produce multiple items or just pass an item through (specifically I am using redux-observable and trying to produce multiple actions based on an api return if it matters).
I have a simplified example below but it doesn't feel like idiomatic rx and just feels weird. In the example if the value is even I want to produce two items, but if odd, just pass the value through. What is the "right" way to achieve this?
test('url and response can be flatMap-ed into multiple objects based on array response and their values', async () => {
const fakeUrl = 'url';
axios.request.mockImplementationOnce(() => Promise.resolve({ data: [0, 1, 2] }));
const operation$ = of(fakeUrl).pipe(
mergeMap(url => request(url)),
mergeMap(resp => resp.data),
mergeMap(i =>
merge(
of(i).pipe(map(num => `number was ${num}`)),
of(i).pipe(
filter(num => num % 2 === 0),
map(() => `number was even`)
)
)
)
);
const result = await operation$.pipe(toArray()).toPromise();
expect(result).toHaveLength(5);
expect(axios.request).toHaveBeenCalledTimes(1);
});
Personally I'd do it in a very similar way. You just don't need to be using the inner merge for both cases:
...
mergeMap(i => {
const source = of(`number was ${i}`);
return i % 2 === 0 ? merge(source, of(`number was even`)) : source;
})
I'm using concat to append a value after source Observable completes. Btw, in future RxJS versions there'll be endWith operator that will make it more obvious. https://github.com/ReactiveX/rxjs/pull/3679
Try to use such combo - partition + merge.
Here is an example (just a scratch)
const target$ = Observable.of('single value');
const [streamOne$, streamTwo$] = target$.partition((v) => v === 'single value');
// some actions with your streams - mapping/filtering etc.
const result$ = Observable.merge(streamOne$, streamTwo$)';

Maintaining Subject emission order when invoking next within subscription

I'm running into a bug and I've determined it's due to the fact that Subjects when next()ed will fire their events synchronously.
The following code produces the following ouput:
mySubject.subscribe(score => {
if (score === 2) {
mySubject.next(score + 10);
}
})
mySubject.subscribe(score => console.log(score))
Ouput:
1
12
2
The only way Im able to get the proper output (1,2,12) is if I wrap my next() call in a setTimeout to make it async. Is there a proper way to deal with this issue that I'm missing?
If you're using RxJS 5.5 I'd personally use setTimeout as well. There's subscribeOn operator that you could use with the async scheduler (import { async } from 'rxjs/scheduler/async') to run every emission in a new frame but it's not available in RxJS 5.5 right now.
So probably the easiest way is using delay(0) which doesn't make any delay and passes everything asynchronously like you did with setTimeout():
import { Subject } from 'rxjs/Subject';
import { delay } from 'rxjs/operators';
const mySubject = new Subject();
const source = mySubject.pipe(delay(0));
source.subscribe(score => {
if (score === 2) {
mySubject.next(score + 10);
}
})
source.subscribe(score => console.log(score));
mySubject.next(1);
mySubject.next(2);
See live demo (open console): https://stackblitz.com/edit/typescript-fiwgrk?file=index.ts
How about this?
const plusTen$ = mySubject.filter(score => score === 2).map(score => score + 10);
mySubject.merge(plusTen$).subscribe(score => console.log(score))
Don't know if it helps the OP, but to anyone who encounters this situation:
It's strictly speaking not a bug because when you next() a Subject, all subscribers are iterated and called. So if you next() within the first subscriber, the iteration starts again and the second subscriber of the first iteration is called after the second iteration is done.
Note that because of this, the order of subscription is relevant. So alternatively to using setTimout or delay as suggested in the accepted answer, you could also swap the subscriptions like this:
mySubject.subscribe(score => console.log(score));
mySubject.subscribe(score => {
if (score === 2) {
mySubject.next(score + 10);
}
})
Output:
1
2
12

Have withLatestFrom wait until all sources have produced one value

I'm making use of the withLatestFrom operator in RxJS in the normal way:
var combined = source1.withLatestFrom(source2, source3);
...to actively collect the most recent emission from source2 and source3 and to emit all three value only when source1 emits.
But I cannot guarantee that source2 or source3 will have produced values before source1 produces a value. Instead I need to wait until all three sources produce at least one value each before letting withLatestFrom do its thing.
The contract needs to be: if source1 emits then combined will always eventually emit when the other sources finally produce. If source1 emits multiple times while waiting for the other sources we can use the latest value and discard the previous values. Edit: as a marble diagram:
--1------------2---- (source)
----a-----b--------- (other1)
------x-----y------- (other2)
------1ax------2by--
--1------------2---- (source)
------a---b--------- (other1)
--x---------y------- (other2)
------1ax------2by--
------1--------2---- (source)
----a-----b--------- (other1)
--x---------y------- (other2)
------1ax------2by--
I can make a custom operator for this, but I want to make sure I'm not missing an obvious way to do this using the vanilla operators. It feels almost like I want combineLatest for the initial emit and then to switch to withLatestFrom from then on but I haven't been able to figure out how to do that.
Edit: Full code example from final solution:
var Dispatcher = new Rx.Subject();
var source1 = Dispatcher.filter(x => x === 'foo');
var source2 = Dispatcher.filter(x => x === 'bar');
var source3 = Dispatcher.filter(x => x === 'baz');
var combined = source1.publish(function(s1) {
return source2.publish(function(s2) {
return source3.publish(function(s3) {
var cL = s1.combineLatest(s2, s3).take(1).do(() => console.log('cL'));
var wLF = s1.skip(1).withLatestFrom(s2, s3).do(() => console.log('wLF'));
return Rx.Observable.merge(cL, wLF);
});
});
});
var sub1 = combined.subscribe(x => console.log('x', x));
// These can arrive in any order
// and we can get multiple values from any one.
Dispatcher.onNext('foo');
Dispatcher.onNext('bar');
Dispatcher.onNext('foo');
Dispatcher.onNext('baz');
// combineLatest triggers once we have all values.
// cL
// x ["foo", "bar", "baz"]
// withLatestFrom takes over from there.
Dispatcher.onNext('foo');
Dispatcher.onNext('bar');
Dispatcher.onNext('foo');
// wLF
// x ["foo", "bar", "baz"]
// wLF
// x ["foo", "bar", "baz"]
I think the answer is more or less as you described, let the first value be a combineLatest, then switch to withLatestFrom. My JS is hazy, but I think it would look something like this:
var selector = function(x,y,z) {};
var combined = Rx.Observable.concat(
source1.combineLatest(source2, source3, selector).take(1),
source1.withLatestFrom(source2, source3, selector)
);
You should probably use publish to avoid multiple subscriptions, so that would look like this:
var combined = source1.publish(function(s1)
{
return source2.publish(function(s2)
{
return source3.publish(function(s3)
{
return Rx.Observable.concat(
s1.combineLatest(s2, s3, selector).take(1),
s1.withLatestFrom(s2, s3, selector)
);
});
});
});
or using arrow functions...
var combined = source1.publish(s1 => source2.publish(s2 => source3.publish(s3 =>
Rx.Observable.concat(
s1.combineLatest(s2, s3, selector).take(1),
s1.withLatestFrom(s2, s3, selector)
)
)));
EDIT:
I see the problem with concat, the withLatestFrom isn't getting the values. I think the following would work:
var combined = source1.publish(s1 => source2.publish(s2 => source3.publish(s3 =>
Rx.Observable.merge(
s1.combineLatest(s2, s3, selector).take(1),
s1.skip(1).withLatestFrom(s2, s3, selector)
)
)));
...so take one value using combineLatest, then get the rest using withLatestFrom.
I wasn't quite satisfied with the accepted answer, so I ended up finding another solution. Many ways to skin a cat!
My use-case involves just two streams - a "requests" stream and a "tokens" stream. I want requests to fire as soon as they are received, using the whatever the latest token is. If there is no token yet, then it should wait until the first token appears, and then fire off all the pending requests.
I wasn't quite satisfied with the accepted answer, so I ended up finding another solution. Essentially I split the request stream into two parts - before and after first token arrives. I buffer the first part, and then re-release everything in one go once I know that the token stream is non-empty.
const first = token$.first()
Rx.Observable.merge(
request$.buffer(first).mergeAll(),
request$.skipUntil(first)
)
.withLatestFrom(token$)
See it live here: https://rxviz.com/v/VOK2GEoX
For RxJs 7:
const first = token$.first()
merge(
request$.pipe(
buffer(first),
mergeAll()
),
request$.pipe(
skipUntil(first)
)
).pipe(
withLatestFrom(token$)
)
I had similar requirements but for just 2 observables.
I ended up using switchMap+first:
observable1
.switchMap(() => observable2.first(), (a, b) => [a, b])
.subscribe(([a, b]) => {...}));
So it:
waits until both observables emit some value
pulls the value from second observable only if the first one has changed (unlike combineLatest)
doesn't hang subscribed on second observable (because of .first())
In my case, second observable is a ReplaySubject. I'm not sure if it will work with other observable types.
I think that:
flatMap would probably work too
it might be possible to extend this approach to handle more than 2 observables
I was surprised that withLatestFrom will not wait on second observable.
In my mind, the most elegant way to achieve the different behavior of an existing RxJS operator is to wrap it into a custom operator. So that from the outside it looks just like any regular operator and doesn't require you to restructure your code each time you need this behavior.
Here is how you can create your own operator which behaves just like withLatestFrom, except that at the very beginning it will emit as soon as the first value of the target observable is emitted (unlike standard withLatestFrom, which will ignore the first emission of the source if the target hasn't yet emitted once). Let's call it delayedWithLatestFrom.
Note that it's written in TypeScript, but you can easily transform it to plain JS. Also, it's a simple version that supports only one target observable and no selector function - you can extend it as needed from here.
export function delayedWithLatestFrom<T, N>(
target$: Observable<N>
): OperatorFunction<T, [T, N]> {
// special value to avoid accidental match with values that could originate from target$
const uniqueSymbol = Symbol('withLatestFromIgnore');
return pipe(
// emit as soon target observable emits the first value
combineLatestWith<T, [N]>(target$.pipe(first())),
// skip the first emission because it's handled above, and then continue like a normal `withLatestFrom` operator
withLatestFrom(target$.pipe(skip(1), startWith(uniqueSymbol))),
map(([[rest, combineLatestValue], withLatestValue]) => {
// take combineLatestValue for the first time, and then always take withLatestValue
const appendedValue =
withLatestValue === uniqueSymbol ? combineLatestValue : withLatestValue;
return [rest, appendedValue];
})
);
}
// SAMPLE USAGE
source$.pipe(
delayedWithLatestFrom(target$)
).subscribe(console.log);
So if you compare it with the original marble diagram for withLatestFrom, it will differ only in one fact: while withLatestFrom ignores the first emissions and produces b1 as the first value, the delayedWithlatestFrom operator will emit one more value a1 at the beginning, as soon as the second observable emits 1.
a) Standard withLatestFrom:
b) Custom delayedWithLatestFrom:
Use combineLatest and filter to remove tuples before first full set is found then set a variable to stop filtering. The variable can be within the scope of a wrapping defer to do things properly (support resubscription). Here it is in java (but the same operators exist in RxJs):
Observable.defer(
boolean emittedOne = false;
return Observable.combineLatest(s1, s2, s3, selector)
.filter(x -> {
if (emittedOne)
return true;
else {
if (hasAll(x)) {
emittedOne = true;
return true;
} else
return false;
}
});
)
I wanted a version where tokens are fetched regularly - and where I want to retry the main data post on (network) failure. I found shareReplay to be the key. The first mergeWith creates a "muted" stream, which causes the first token to be fetched immediately, not when the first action arrives. In the unlikely event that the first token will still not be available in time, the logic also has a startWith with an invalid value. This causes the retry logic to pause and try again. (Some/map is just a Maybe-monad):
Some(fetchToken$.pipe(shareReplay({refCount: false, bufferSize: 1})))
.map(fetchToken$ =>
actions$.pipe(
// This line is just for starting the loadToken loop immediately, not waiting until first write arrives.
mergeWith(fetchToken$.pipe(map(() => true), catchError(() => of(false)), tap(x => loggers.info(`New token received, success: ${x}`)), mergeMap(() => of()))),
concatMap(action =>
of(action).pipe(
withLatestFrom(fetchToken$.pipe(startWith(""))),
mergeMap(([x, token]) => (!token ? throwError(() => "Token not ready") : of([x, token] as const))),
mergeMap(([{sessionId, visitId, events, eventIds}, token]) => writer(sessionId, visitId, events, token).pipe(map(() => <ISessionEventIdPair>{sessionId, eventIds}))),
retryWhen(errors =>
errors.pipe(
tap(err => loggers.warn(`Error writing data to WG; ${err?.message || err}`)),
mergeMap((_error: any, attemptIdx) => (attemptIdx >= retryPolicy.retryCount ? throwError(() => Error("It's enough now, already")) : of(attemptIdx))), // error?.response?.status (int, response code) error.code === "ENOTFOUND" / isAxiosError: true / response === undefined
delayWhen(attempt => timer(attempt < 2 ? retryPolicy.shortRetry : retryPolicy.longRetry, scheduler))
)
)
)
),
)
)
Thanks to everyone on this question-page for good inputs.
Based on the answer from #cjol
Here's a RxJs 7 implementation of a waitFor operator that will buffer the source stream until all input observables have emitted values, then emit all buffered events on the source stream. Any subsequent events on the source stream are emitted immediately.
// Copied from the definition of withLatestFrom() operator.
export function waitFor<T, O extends unknown[]>(
inputs: [...ObservableInputTuple<O>]
): OperatorFunction<T, [T, ...O]>;
/**
* Buffers the source until every observable in "from" have emitted a value. Then
* emit all buffered source values with the latest values of the "from" array.
* Any source events are emitted immediately after that.
* #param from Array of observables to wait for.
* #returns Observable that emits an array that concatenates the source and the observables to wait.
*/
export function waitFor(
from: Observable<unknown>[]
): (source$: Observable<unknown>) => Observable<unknown> {
const combined$ = combineLatest(from);
// This served as a conditional that switched on and off the streams that
// wait for the the other observables, or emits the source right away because
// the other observables have emitted.
const firstCombined$ = combined$.pipe(first());
return function (source$: Observable<unknown>): Observable<unknown> {
return merge(
// This stream will buffer the source until the other observables have all emitted.
source$.pipe(
takeUntil(firstCombined$), // without this it continues to buffer new values forever
buffer(firstCombined$),
mergeAll()
),
// This stream emits the source straight away and will take over when the other
// observables have emitted.
source$.pipe(skipUntil(firstCombined$))
).pipe(
withLatestFrom(combined$),
// Flatten it to behave like withLatestFrom() operator.
map(([source, combined]) => [source, ...combined])
);
};
}
All of the above solutions are not really on the point, therefore I made my own. Hope it helps someone out.
import {
combineLatest,
take,
map,
ObservableInputTuple,
OperatorFunction,
pipe,
switchMap
} from 'rxjs';
/**
* ### Description
* Works similar to {#link withLatestFrom} with the main difference that it awaits the observables.
* When all observables can emit at least one value, then takes the latest state of all observables and proceeds execution of the pipe.
* Will execute this pipe only once and will only retrigger pipe execution if source observable emits a new value.
*
* ### Example
* ```ts
* import { BehaviorSubject } from 'rxjs';
* import { awaitLatestFrom } from './await-latest-from.ts';
*
* const myNumber$ = new BehaviorSubject<number>(1);
* const myString$ = new BehaviorSubject<string>("Some text.");
* const myBoolean$ = new BehaviorSubject<boolean>(true);
*
* myNumber$.pipe(
* awaitLatestFrom([myString$, myBoolean$])
* ).subscribe(([myNumber, myString, myBoolean]) => {});
* ```
* ### Additional
* #param observables - the observables of which the latest value will be taken when all of them have a value.
* #returns a tuple which contains the source value as well as the values of the observables which are passed as input.
*/
export function awaitLatestFrom<T, O extends unknown[]>(
observables: [...ObservableInputTuple<O>]
): OperatorFunction<T, [T, ...O]> {
return pipe(
switchMap((sourceValue) =>
combineLatest(observables).pipe(
take(1),
map((values) => [sourceValue, ...values] as unknown as [T, ...O])
)
)
);
}
Actually withLatestFrom already
waits for every source
emits only when source1 emits
remembers only the last source1-message while the other sources are yet to start
// when source 1 emits the others have emitted already
var source1 = Rx.Observable.interval(500).take(7)
var source2 = Rx.Observable.interval(100, 300).take(10)
var source3 = Rx.Observable.interval(200).take(10)
var selector = (a,b,c) => [a,b,c]
source1
.withLatestFrom(source2, source3, selector)
.subscribe()
vs
// source1 emits first, withLatestFrom discards 1 value from source1
var source1 = Rx.Observable.interval(500).take(7)
var source2 = Rx.Observable.interval(1000, 300).take(10)
var source3 = Rx.Observable.interval(2000).take(10)
var selector = (a,b,c) => [a,b,c]
source1
.withLatestFrom(source2, source3, selector)
.subscribe()

Resources