如何將源轉換為流？
輸入：Source[ByteString,NotUsed] 中間步驟：呼叫回傳 InputStream 的 API 輸出：Flow[ByteString,ByteString,NotUsed]
我這樣做是：輸入型別是 = Source[ByteString,NotUsed]
val sink: Sink[ByteString,InputStream] = StreamConverters.asInputStream()
val output: InputStream = <API CALL>
val mySource: Source[ByteString,Future[IOResult]] = StreamConverters.fromInputStream(() => output)
val myFlow: Flow[ByteString,ByteString,NotUsed] = Flow.fromSinkAndSource(sink,source)
當我在源中使用上述 Flow 時，它回傳一個空結果。有人可以幫我弄清楚我做得對嗎？
uj5u.com熱心網友回復：
我不確定您是否完全掌握您想要實作的目標，但也許這是一個用例flatMapConcat：
def readInputstream(bs: ByteString): Source[ByteString, Future[IOResult]] =
// Get some IS from the ByteString
StreamConverters.fromInputStream(() => ???)
val myFlow: Flow[ByteString, ByteString, NotUsed] =
Flow.flatMapConcat(bs => readInputstream(bs))
// And use it like this:
val source: Source[ByteString] = ???
source
.via(myFlow)
.to(???)
轉載請註明出處，本文鏈接：https://www.uj5u.com/gongcheng/524176.html
標籤：斯卡拉阿卡流
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