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scala从REST服务获取响应

playframework scala

0

Udders · 技术社区 · 7 年前

如果200,则将响应数据推到视图中并显示它。

我写了一个这样的连接器,

package connectors

import com.google.inject.Inject
import config.AppConfig
import play.api.libs.json.Json
import play.api.libs.ws.{WSClient, WSResponse}
import play.mvc.Http.Status
import scala.concurrent.ExecutionContext.Implicits.global
import scala.concurrent.Future
import play.api.libs.json._
class RegisterConnector @Inject() (appConfig: AppConfig, ws: WSClient) {

  val customerBaseUrl: String = appConfig.customerBaseUrl

  def verifyCustomer(code: String): Future[Boolean] = {
    ws.url(s"$customerBaseUrl/customer/$code").post(code).map(processResponseStatus)
  }

  private def processResponseStatus(resp: WSResponse): Boolean = resp.status match {
    case code if (Status.OK until Status.MULTIPLE_CHOICES).contains(code) => true
    case _ => throw new RegisterException(s"Server error: ${resp.status} . ${resp.body}");
  }

}

class RegisterException(message: String) extends RuntimeException(message: String)

    package controllers

import com.google.inject.{Inject, Singleton}
import config.AppConfig
import play.api.i18n.{I18nSupport, MessagesApi}
import play.api.mvc._
import scala.concurrent._
import ExecutionContext.Implicits.global
import connectors.RegisterConnector
import play.api.Logger
import views.html.register.verified

@Singleton
class VerifiedController @Inject()(registerConnector: RegisterConnector,
                                    val messagesApi: MessagesApi)
                                   (implicit appConfig: AppConfig) extends Controller with I18nSupport {

  def verify(code: String): Action[AnyContent] = Action.async { implicit request: Request[AnyContent] =>

    registerConnector.verifyCustomer(code).map {
      case true =>

      case _ =>

    }.recover {
      case ex =>

    }

  }

  def show(): Action[AnyContent] = Action { implicit request: Request[AnyContent] =>
    Ok(verified())
  }

  def error(): Action[AnyContent] = Action { implicit request: Request[AnyContent] =>
    Ok(verified())
  }

}

一些帮助和指导将是非常棒的!

1 回复 | 直到 7 年前

1

Bruno Paulino 7 年前

假设JSON是响应格式,我通常开始在case类中包装响应结构。假设以下JSON响应:

{
  "city": "Vienna",
  "forecast": 10
}

现在这个案子看起来是这样的:

case class WeatherUpdate(city: String, forecast: Int)

使用singleton进行API调用的策略非常好。我们给他打电话 WeatherService Future[Boolean] 您可以使用Scala标准库 Try .

class WeatherService @Inject() (ws: WSClient) {

  private val logger = Logger(getClass)

  def getWeather(): Future[Try[WeatherUpdate]] = {
    ws.get("https://myweather.com/api/update") // this is all fake
      .map {
        case response if response.status == 200 =>
          val city = (response.json \ "city").as[String]
          val forecast = (response.json \ "forecast").as[Int]
          Success(WeatherUpdate(city, forecast))
        case response =>
          Failure(new WeatherException("Could not get weather response"))
      }.recover { // always recover from a future. It could throw an exception and you would never know
        case ex: Exception =>
          logger.error("could not get weather info", ex)
          Failure(ex)
      }
  }

}

class WeatherException(message: String) extends RuntimeException(message: String)

在控制器上,现在可以使用天气更新渲染模板:

def verify(code: String): Action[AnyContent] = Action.async { 
  implicit request =>
    weatherService.getWeather().map {
      case Success(weatherUpdate) => Ok(weather_template(weatherUpdate))
      case Failure(ex) => BadRequest(weather_error())
    }
}