0

我对 kotlin 和 rxjava 有意外行为。我创建了一个使用毕加索加载图像的扩展函数

fun Picasso.loadBitmap(url: String) : Observable<Bitmap>
        = Observable.create<Bitmap> {
    emitter ->
    Log.d("picasso load bitmap", "me ${this}")
    try {
        val bitmap = load(url).centerCrop()
                .resize(100, 100)
                .transform(CircleTransformer())
                .get()
        emitter.onNext(bitmap)
        emitter.onComplete()
    } catch (e: IOException) {
        emitter.onError(e)
    }
}

我像这样在很近的时间间隔内（几乎同时）多次调用它，

picasso.loadBitmap(place.image_url)
    .subscribeOn(Schedulers.io())
    .retryWhen { error ->
        error.zipWith(Observable.range(1, 5),
                BiFunction<Throwable, Int, RetryWrapper> {
                    t1, t2 -> RetryWrapper(t2.toLong(), t1) })
                .flatMap {
                    if(it.delay < 4){
                        Log.d(TAG, "retry no. ${it.delay} for ${place.image_url}")
                        Observable.timer(it.delay * 5, TimeUnit.SECONDS)
                    } else {
                        Log.d(TAG, "DMD ${place.image_url}")
                        Observable.error { it.error }
                    }
                }

    }
    .subscribe (
        { bitmap ->
            markers.find { it.place.id == place.id }?.let {
                it.marker.icon = IconFactory.getInstance(context).fromBitmap(bitmap)
            }
        },
        {
            Log.e(TAG, "error decoding ${place.image_url}", it)
        })

我希望每次loadBitmap都会被调用，它会创建一个新的 observable。但我在日志中得到了这个

09-28 11:17:00.022 31694-32276/? D/picasso load bitmap: me com.squareup.picasso.Picasso@c894e26
09-28 11:17:00.068 31694-32277/? D/picasso load bitmap: me com.squareup.picasso.Picasso@c894e26
09-28 11:17:00.069 31694-31959/? D/picasso load bitmap: me com.squareup.picasso.Picasso@c894e26
09-28 11:17:00.108 31694-32278/? D/picasso load bitmap: me com.squareup.picasso.Picasso@c894e26
09-28 11:17:00.112 31694-32251/? D/picasso load bitmap: me com.squareup.picasso.Picasso@c894e26
09-28 11:17:00.125 31694-32260/? D/picasso load bitmap: me com.squareup.picasso.Picasso@c894e26
09-28 11:17:00.162 31694-31794/? D/picasso load bitmap: me com.squareup.picasso.Picasso@c894e26
09-28 11:17:00.192 31694-32280/? D/picasso load bitmap: me com.squareup.picasso.Picasso@c894e26
09-28 11:17:00.195 31694-32279/? D/picasso load bitmap: me com.squareup.picasso.Picasso@c894e26
09-28 11:17:00.219 31694-32281/? D/picasso load bitmap: me com.squareup.picasso.Picasso@c894e26
09-28 11:17:04.828 31694-32262/? D/picasso load bitmap: me com.squareup.picasso.Picasso@c894e26
09-28 11:17:14.885 31694-31793/? D/picasso load bitmap: me com.squareup.picasso.Picasso@c894e26
09-28 11:17:29.928 31694-32269/? D/picasso load bitmap: me com.squareup.picasso.Picasso@c894e26

loadBitmap所有调用的 observable 都是相同的。我需要他们有自己的 observable，因为如果我不这样做，当retryWhen失败时，它不会继续下一次失败。我希望它有意义。

将 observable 放入 adefer或flatmap不会改变任何东西。

编辑我的代码

override fun render(state: MainState) {
        map?.let { map ->
            val newMarkers: MutableList<PlaceMarker> = mutableListOf()
            for(place in state.places) {
                var placeMarker = placeMarkers.find { it.place.id == place.id }
                if(placeMarker != null && map.markers.contains(placeMarker.marker)) {
                    newMarkers.add(placeMarker)
                    placeMarkers.remove(placeMarker)
                } else {
                    if(placeMarker != null) placeMarkers.remove(placeMarker)
                    val option = MarkerOptions()
                    option.position = LatLng(place.latitude, place.longitude)
                    option.snippet = place.name
                    placeMarker = PlaceMarker(place, map.addMarker(option))
                    newMarkers.add(placeMarker)

                    picasso.loadBitmap(place.image_url)
                            .subscribeOn(Schedulers.io())
                            .retryWhen { error ->
                                error.zipWith(Observable.range(1, 5),
                                        BiFunction<Throwable, Int, RetryWrapper> {
                                            t1, t2 -> RetryWrapper(t2.toLong(), t1) })
                                        .flatMap {
                                            if(it.delay < 4){
                                                Log.d(TAG, "retry no. ${it.delay} for ${place.image_url}")
                                                Observable.timer(it.delay * 5, TimeUnit.SECONDS)
                                            } else {
                                                Log.d(TAG, "DMD ${place.image_url}")
                                                Observable.error { it.error }
                                            }
                                        }
                            }
                            .subscribe (
                                    { bitmap ->
                                        placeMarkers.find { it.place.id == place.id }?.let {
                                            it.marker.icon = IconFactory.getInstance(context).fromBitmap(bitmap)
                                            bitmap.recycle()
                                        }
                                    },
                                    {
                                        Log.e(TAG, "error decoding ${place.image_url}", it)
                                    })
                }

            }
            placeMarkers.forEach { it.marker.remove() }
            placeMarkers.clear()
            placeMarkers.addAll(newMarkers)
        }
    }

我使用 MVP，只是为了让您看到更广泛的内容。所以，这是一个视图中的函数，渲染将在模型完成从服务器获取数据后触发。

4

1 回答 1

3

你必须在这里小心。关键字this在

Log.d("picasso load bitmap", "me ${this}")

不针对Observable但接收者类型。在你的情况下Picasso。你在你的日志中看到me com.squareup.picasso.Picasso@c894e26

好消息是，Observable每次调用loadBitmap. 您可以通过以下方式检查：

val observable = picasso.loadBitmap(place.image_url)
Log.d("observable for picasso", "$observable")
observable.subscribeOn(Schedulers.io())...

所以你看，你loadBitmap总是在同一个实例上调用，picasso这就是你为那个类获得相同输出的原因。但是每个人的呼唤loadBitmap都会创造一个新的Observable

所以你的代码很好。

于 2017-09-28T06:16:55.483 回答