RxJava：如何拆分单

Question

如何将单个流拆分为单独的单个流，这样就可以执行以下操作而无需两次计算getUserId()？

// getUserId() returns Single<String>

getUserId().flatMap { getSomething(it) }  // Return Single<Something>
getUserId().flatMap { getSomethingElse(it) } // Return Single<SomethingElse>

Answer 1

使用getUserId缓存cache的结果

val userIdCached = getUserId().cache()
userIdCached
    .flatMap { getSomething(it) }
    .subscribe(...)
userIdCached
    .flatMap { getSomethingElse(it) }
    .subscribe(...)