scala – 如何解构Spray API HTTPResponse？

如果我只是期望和HTTPResponse,我得到一个值,但是当我尝试将它编组到我的case类中时,它失败了(我不能生成除null()之外的消息,因为我正在使用匹配和显然没有得到正确的测试用例.)

我认为我的一些问题是它以text / plain的形式返回数据而不是application / json.当我期待HttpResponse而不是我的案例类时,

val f: Future[HttpResponse] =
    (IO(Http) ? Get("http://1.2.3.4:8983/solr/collection1/select?q=*%3A*&wt=json")).mapTo[HttpResponse]

我明白了：

HttpResponse(200 OK,HttpEntity(text/plain; charset=UTF-8,{
  "responseHeader":{"status":0,"QTime":65,"params":{"q":"*:*","wt":"json"}},"response":{"numFound":147437873,"start":0,"maxScore":1.0,"docs":
    [
      {"guid":"TLQ0jVlMYCXQrYkBIZHNXfMmifw+3","alias":["greg"],"_version_":1440942010264453120},{"guid":"TQsDY1ZG7q+Ne5e6F7qAUhFyomSH9","_version_":1440942010296958976},{"guid":"TzWB5grOBAJJZcAQDo2k9xBUVGPFr","alias":["spark"],"_version_":1440942010298007552},{"guid":"T0judCG4UI9RYqDDQVcn+gyZEU7Bb","alias":["zombie"],...),List(Connection: close,Content-Type: text/plain; charset=UTF-8),HTTP/1.1)

但是当我改变它以期待我的案例类时,我无法比拟.那么,我如何将它返回的数据编组到Scala案例类中？这是我尝试过的：

case class SolrParams(q: String,wt: String)
case class SolrResponseHeader(status: String,qtime: String,params: SolrParams)
case class SolrDoc(guid: String,alias: List[String],version: String)
case class SolrResponse(numFound: Long,start: Long,maxScore: String,docs: List[SolrDoc])

case class SolrApResult(responseHeader: SolrResponseHeader,response: SolrResponse)

object SolrJsonProtocol extends DefaultJsonProtocol {
  implicit val paramsFormat = jsonFormat2(SolrParams)
  implicit val responseHeaderFormat = jsonFormat2(SolrResponseHeader)
  implicit val docFormat = jsonFormat3(SolrDoc)
  implicit val responseFormat = jsonFormat4(SolrResponse)
  implicit def solrApiResultFormat = jsonFormat2(SolrApiFullResult)
}

...

val f: Future[SolrApiResult] =
    (IO(Http) ? Get("http://1.2.3.4:8983/solr/collection1/select?q=*%3A*&wt=json")).mapTo[SolrApiResult]

这使我在f onComplete …结构中无法匹配.问题可能是我的案例类与返回的内容不匹配,如果是,那么你有什么建议可以更好地解决它？

我已经all over the docs了,他们要么不完整还是有点过时,加上我在这场比赛中是新手,所以这也没有帮助.