Better way to chain rxjava2 calls with conditional operations

Question

I have the following code, that does one single call, gets the result of the call, which is a boolean, then makes the second call if the result is false.

private fun linkEmailAndTextTogether(contactPhoneNumber: ContactPhoneNumbers,phoneNumber : PhoneNumber) {
    val single = SingleOnSubscribe<Boolean> {
        contactPhoneNumber.doesEmailContactExist(phoneNumber)
    }
    Single.create(single)
            .subscribeOn(Schedulers.io())
            .subscribeWith(object : SingleObserver<Boolean> {
                override fun onSuccess(phoneNumberDoesExist: Boolean) {
                    if (!phoneNumberDoesExist) {
                        val completable = CompletableOnSubscribe {
                            contactPhoneNumber.linkEmailAndTextTogether(phoneNumber)
                        }
                        compositeDisposable.add(Completable.create(completable)
                                .subscribeOn(Schedulers.io())
                                .subscribe())
                    }
                }

                override fun onSubscribe(d: Disposable) {
                    compositeDisposable.add(d)
                }

                override fun onError(e: Throwable) {
                    Timber.e(e,e.localizedMessage)
                }


            })
}

It seems like there should be a more elegant way to do this in some kind of chain.

Maybe similar: https://stackoverflow.com/questions/36058320/rxjava-how-to-conditionally-apply-operators-to-an-observable-without-breaking-t — Christophe Roussy, Feb 15 '18 at 15:58
That didn't really help me understand what I should do. But thanks. — Kristy Welsh, Feb 15 '18 at 16:01
https://stackoverflow.com/questions/39526978/what-rxjava-operator-to-use-to-chain-observables-only-under-certain-conditions?rq=1 this should do. — karandeep singh, Feb 15 '18 at 16:05
@Blackbelt - I'm new to RX java and don't exactly know how to use the map operators — Kristy Welsh, Feb 15 '18 at 16:07
@karandeepsingh I also have single and a completable - how to chain those? — Kristy Welsh, Feb 15 '18 at 16:10
Good tutorial here: https://github.com/Froussios/Intro-To-RxJava, also learning this stuff. — Christophe Roussy, Feb 15 '18 at 16:14

score 1 · Accepted Answer · answered Feb 15 '18 at 16:23

you could use the flatMap operator - the downside is that you won't know if the first or the second failed.

  Single.just(phoneNumber)
       .subscribeOn(Schedulers.io())
       .map { it -> contactPhoneNumber.doesEmailContactExist(it) }
       .flatMap { it ->
             if (it) {
                 return@flatMap contactPhoneNumber.linkEmailAndTextTogether(phoneNumber)
              }
              Single.just(it)
        }.subscribe({}, Throwable::printStackTrace);

score 1 · Answer 2 · answered Feb 15 '18 at 16:24

1

This should help.

val single = SingleOnSubscribe<Boolean> {
     getSingle()
   }

   Single.create(single).map({
    if (it){
        return@map getCompleteable()
    }
    return@map Completable.complete()
})

answered Feb 15 '18 at 16:24

karandeep singh

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Better way to chain rxjava2 calls with conditional operations

2 Answers2