Akka http building Uri

0
votes

I'm trying to build Uri, but Query method gives me type mismatch, expected string, actual Uri.Query

def buildUri(url: String, query: Option[Map[String, String]] = None): Uri = {

    val fullUrl = endpoint + (if (endpoint.endsWith("/")) url else "/" + url)

    val uri: Uri = query match {
      case Some(map) if map.nonEmpty =>
        Uri(fullUrl.toString).copy(query = Query(map))
      case _ =>
        Uri(fullUrl.toString)
    }

    uri

  }
1
scalaakkaakka-http
Uri case class doesn't has any parameter named query as i can see. Uri(scheme: String, authority: Authority, path: Path, rawQueryString: Option[String], fragment: Option[String]) – Raman Mishra
There is object Queri inside – ifog
Query not query and you are assiging value to query it doesn't has any parameter named query and yes it has Query. what you exactly want to do? – Raman Mishra

1 Answers

1
votes

I think this is what you want to do:

def buildUri(url: String, query: Option[Map[String, String]] = None): Uri = {

    val fullUrl = endpoint + (if (endpoint.endsWith("/")) url else "/" + url)

    val uri: Uri = query match {
      case Some(map) if map.nonEmpty =>
        Uri(fullUrl.toString).withQuery(query = Query(map))
      case _ =>
        Uri(fullUrl.toString)
    }

    uri
  }