scala返回Future[Unit]而不是Future[ContentComponentModel]

Question

我有以下代码：

  def getContentComponents: Action[AnyContent] = Action.async {
        val test = contentComponentDTO.list().map { contentComponentsFuture =>
          contentComponentsFuture.foreach(contentComponentFuture =>

            contentComponentFuture.typeOf match {
              case 1 =>
                println("blubb")
              case 5 =>
                contentComponentDTO.getContentComponentText(contentComponentFuture.id.get).map(
                  text => {
                    contentComponentFuture.text = text.text
                    println(text.text)
                    println(contentComponentFuture.text)
                  }
                )
            }
          )

        }

    Future.successful(Ok(Json.obj("contentComponents" -> test)))

  }

我收到了一条错误信息：

​

​

.list()方法应该返回一个FutureContentComponentModel

def list(): Future[Seq[ContentComponentModel]] = db.run {

我在这个案子里犯了什么错误？

谢谢

Answer 1

您的contentComponentsFuture应该是SeqContentComponentModel类型的。在这种情况下，你应该移动

Future.successful(Ok(Json.obj("contentComponents" -> test)))

只进入映射表达式(这是异步的)后循环。

它应该看起来像：

def getContentComponents: Action[AnyContent] = Action.async {
val test = contentComponentDTO.list().map { contentComponents =>
  contentComponents.foreach(contentComponentFuture =>
    contentComponentFuture.typeOf match {
      case 1 =>
        println("blubb")
      case 5 =>
        contentComponentDTO.getContentComponentText(contentComponentFuture.id.get).map(
          text => {
            contentComponentFuture.text = text.text
            println(text.text)
            println(contentComponentFuture.text)
          }
        )
    }
  )
  Future.successful(Ok(Json.obj("contentComponents" -> contentComponents)))
}

}