Question

我刚开始使用Play框架和Scala。

要习惯它我正在开发一个歌词网络应用程序，但我在从数据库中提取记录时遇到一些麻烦...

以下是我的 MusicController.scala 。我在def show(id: Long)收到错误。我调整了我找到的代码here。

我想我需要实施findById(id)，对吗？但是哪里？怎么样？

我最近一直在使用 Laravel ，我发现很难找到Play with Scala的资源和代码示例。在这种情况下，文档具有我需要的内容，但我对于findById(id)的实现位置和方式一无所知。我错过了什么吗？

Music.scala

package models

import play.api.libs.json.Json

case class Music(id: Long, title: String, lyrics: String, year: Int)

object Music {
  implicit val musicFormat = Json.format[Music]
}

MusicController.scala

package controllers

import javax.inject.Inject

import dal.MusicRepository
import models.Music
import play.api.data.Form
import play.api.data.Forms._
import play.api.data.validation.Constraints._
import play.api.i18n.{I18nSupport, MessagesApi}
import play.api.libs.json.Json
import play.api.mvc._

import scala.concurrent.{Future, ExecutionContext}

class MusicController @Inject()(repo: MusicRepository, val messagesApi: MessagesApi)
                               (implicit ec: ExecutionContext) extends Controller with I18nSupport {

  def index = Action {
    Ok(views.html.musics.index(musicForm))
  }

  def show(id: Long) = Action {
    Music.findById(id).map { music =>
      Ok(views.html.musics.show(music))
    }.getOrElse(NotFound)
  }

  val musicForm: Form[CreateMusicForm] = Form {
    mapping(
      "title" -> nonEmptyText,
      "lyrics" -> nonEmptyText,
      "year" -> number.verifying(min(0))
    )(CreateMusicForm.apply)(CreateMusicForm.unapply)
  }

  def addMusic = Action.async { implicit request =>
    musicForm.bindFromRequest.fold(
      errorForm => {
        Future.successful(Ok(views.html.musics.index(errorForm)))
      },
      music => {
        repo.create(music.title, music.lyrics, music.year).map { _ =>
          Redirect(routes.MusicController.index)
        }
      }
    )
  }

  def getMusics = Action.async {
    repo.list().map { musics =>
      Ok(Json.toJson(musics))
    }
  }

}

case class CreateMusicForm(title: String, lyrics: String, year: Int)

MusicRepository.scala

package dal

import javax.inject.Inject

import models.Music
import play.api.db.slick.DatabaseConfigProvider
import slick.driver.JdbcProfile

import scala.concurrent.{Future, ExecutionContext}

class MusicRepository @Inject()(dbConfigProvider: DatabaseConfigProvider)
                               (implicit ec: ExecutionContext) {

  private val dbConfig = dbConfigProvider.get[JdbcProfile]

  import dbConfig._
  import driver.api._

  private class MusicsTable(tag: Tag) extends Table[Music](tag, "musics") {
    def id = column[Long]("id", O.PrimaryKey, O.AutoInc)

    def title = column[String]("title")

    def lyrics = column[String]("lyrics")

    def year = column[Int]("year")

    def * = (id, title, lyrics, year) <>((Music.apply _).tupled, Music.unapply)
  }

  private val musics = TableQuery[MusicsTable]

  def create(title: String, lyrics: String, year: Int): Future[Music] = db.run {
    (musics.map(m => (m.title, m.lyrics, m.year))
      returning musics.map(_.id)

      into ((stuff, id) => Music(id, stuff._1, stuff._2, stuff._3))

      ) +=(title, lyrics, year)
  }

  def list(): Future[Seq[Music]] = db.run {
    musics.result
  }

}

提前谢谢！

Answer 1

我也设法解决了这个问题：

<强> MusicController.scala

  def show(id: Long) = Action.async { implicit request =>
    repo.findById(id).map { music =>
      Ok(views.html.musics.show(music))
    }
  }

<强> MusicRepository.scala

  def findById(id: Long): Future[Music] = db.run {
    musics.filter(_.id === id).result.head
  }

如何使用Play with Scala和Slick

1 个答案: