Scala Akka HighLevelHttp从恢复未来返回路由
因为我有一个请求主体,所以我需要将其作为将来的请求,并在操作之后发送响应。但是,我需要检查主体是否有效,如果无效,我想返回一个BADDREQUEST状态 我知道代码太多,所以请查看Scala Akka HighLevelHttp从恢复未来返回路由,scala,routes,akka,future,akka-http,Scala,Routes,Akka,Future,Akka Http,因为我有一个请求主体,所以我需要将其作为将来的请求,并在操作之后发送响应。但是,我需要检查主体是否有效,如果无效,我想返回一个BADDREQUEST状态 我知道代码太多,所以请查看requestFuture,恢复功能中的问题 (pathPrefix("api" / "search") & post & extractRequest) { request => val tokenResult = request.headers.fi
requestFuture
,恢复功能中的问题
(pathPrefix("api" / "search") & post & extractRequest) { request =>
val tokenResult = request.headers.find(x => x.name.toLowerCase == "token")
tokenResult match {
case None => throw new IllegalArgumentException(ServerMessages.TOKEN_NOT_FOUND)
case Some(token) => token.value match {
case this.apiToken => {
val entity = request.entity
val strictEntityFuture = entity.toStrict(2 seconds)
val requestFuture = strictEntityFuture
.map(_.data.utf8String.parseJson
.convertTo[SearchFilters])
requestFuture map { filters =>
// here - I return an Route with data from controller
complete(controller.searchData(filters)
.map(_.toJson.prettyPrint)
.map(toHttpEntity)))
} recover (e => {
// HERE IS THE PROBLEM
// I need to return an Route, but here will be a Future[Route]
complete(400 -> ServerMessages.INVALID_REQUEST_BODY)
}
}
case _ => throw new IllegalArgumentException(ServerMessages.INVALID_TOKEN)
}
}
我需要从未来解压响应,或者使用另一种方式抛出错误
找到了一个使用onComplete
指令的解决方案,但我需要知道我的json是否成功转换,以便抛出自定义错误
onComplete(requestFuture) {
case Success(filters) => searchKpi(filters) ~
pathEndOrSingleSlash {
complete(403 -> ServerMessages.INVALID_METHOD_ARGUMENTS)
}
case Failure(ex) => failWith(ex) // returns 500 Internal Server Error
}
谢谢试试以下方法:
val requestFuture = strictEntityFuture
requestFuture.map { entity =>
entity.data.utf8String match {
// Do your processing here
case "" => RouteResult.Complete(HttpResponse(StatusCode.int2StatusCode(200), entity = HttpEntity("OK")))
}}.recover {
case err: Throwable =>
RouteResult.Complete(HttpResponse(StatusCode.int2StatusCode(500), entity = HttpEntity(err.getMessage)))
}
你可以使用
像在文档中一样定义异常处理程序,并将异常(应该与failWith
引发的异常相同)添加到需要返回的HTTP状态(您提到BadRequest
)
请注意,有两种方法可以附加异常处理程序
从文档:
我已经有了一个
异常句柄
r,但只有两个案例(IllegalArgument
或NoSuchElement
)。由于我扩展了我的处理程序,可以返回BadRequest
响应。但是我在那里做了一些错误的事情,因为我的BadRequestException
被打包到defaultExceptionHandler
路由
您可以执行类似于处理异常(NoTouchElementExceptionHandler.withFallback(tokenNotFoundExceptionHandler)。withFallback(defaultExceptionHandler))的操作。
case class MyBadException() extends RuntimeException
implicit def myExceptionHandler: ExceptionHandler =
ExceptionHandler {
case MyBadException() => complete(HttpResponse(BadRequest)) // You decide the HTTP status to return
case _ => complete(HttpResponse(InternalServerError, entity = "Bad numbers, bad result!!!"))
}
val route = path("test") {
onComplete(getSomething()) {
case Success(v) => complete("Ok")
case Failure(err) => failWith(MyBadException())
}
}
Bring it into implicit scope at the top-level.
Supply it as argument to the handleExceptions directive.
val routes = (
handleExceptions(noSuchElementExceptionHandler) &
handleExceptions(tokenNotFoundExceptionHandler)) {
domainRoutes
}
private val domainRoutes = {
onComplete(future) {
case Success(values) => complete(controller.getdata(values))
case Failure(e) => failWith(BadRequestExceptionHandler(e.getMessage))
}
}