Isvara
Isvara

Reputation: 3463

How to marshal a Future[Source[...]] to an HttpResponse with akka-http?

I have the following method that returns a Future[Source[List[String]]] (the first two lines of a CSV file):

def get(url: String, charset: String, delimiter: Char, quote: Char, escape: Char) = {
    val scanner = CsvParsing.lineScanner(
        delimiter.toByte,
        quote.toByte,
        escape.toByte
    )

    val request = HttpRequest(GET, Uri(url)).withHeaders(`User-Agent`(UserAgent))

    Http(system)
        .singleRequest(request)
        .map { response =>
            response.entity.withoutSizeLimit.dataBytes
                .viaMat(scanner)(Keep.left)
                .map(row =>
                    row.map(bs =>
                        bs.decodeString(charset)
                    )
                )
                .take(2)
        }
}

The returned Future is passed to complete, which marshals it to a JSON array of arrays using:

implicit val jsonStreamingSupport: JsonEntityStreamingSupport = EntityStreamingSupport.json()

However, I'd like to inspect response and return a different HttpResponse if it isn't a 200. It seems like the best way to do this would be to marshal the Future[Source[...]] to an HttpResponse in this method, and then its return type would be HttpResponse either way.

How do I do that? Or is there a better way?

Upvotes: 2

Views: 492

Answers (1)

Isvara
Isvara

Reputation: 3463

Okay, so I got there in the end with a different approach.

Http(system).singleRequest(request)
    .flatMap { response =>
        response.status match {
            case StatusCodes.OK =>
                val compression = CompressionChooser.choose(url, gzip, response)
                response.entity.withoutSizeLimit.dataBytes
                    .via(compression.decoder.decoderFlow)
                    .viaMat(scanner)(Keep.left)
                    .map(_.map(_.decodeString(charset)))
                    .take(2)
                    .runWith(Sink.seq)
                    .map { rows =>
                        val json = Json.toJson(rows)
                        HttpResponse(
                            StatusCodes.OK,
                            entity = HttpEntity(ContentTypes.`application/json`, json.toString)
                        )
                    }

            case _ => Future successful HttpResponse(StatusCodes.BadRequest, entity = "Error")
        }
    }

Upvotes: 1

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