Question

我坚持使用光滑的查询，但不幸的是找不到类似的例子。

配置：

scalaVersion := "2.11.7"
libraryDependencies += "com.typesafe.play" %% "play-slick" % "2.1.0"

在此情况下。我有一个名为Record的表/模型。对象本身包含两个序列，即Tags和Markets。这是数据库结构的代表图像（我知道这不是ER图，也并非意味着）：

Tags和Markets拥有自己的表，并通过多对多关系连接到Record。目标是建立一个查询，以检索所有记录（无论标签和市场如何），具有市场的记录和具有标签的记录。我有这样的想法：

Future[Seq[(RecordModel, Option[Seq[MarketModel]], Option[Seq[TagModel]])]]

这就是我所拥有的：

def myFunction(): Future[Seq[(RecordModel, Seq[MarketModel], Seq[TagModel])]] = {
  val query = for {
    recs <- records joinLeft (recordsMarkets join markets on (_.marketId === _.marketId)) on (_.recordId === _._1.recordId) joinLeft (recordsTags join tags on (_.tagId === _.tagId)) on (_._1.recordId === _._1.recordId)
  } yield recs
  db.run(query.result).map(_.toList.groupBy(_._1).map {
    case (r, m) => (
      r._1, // Records
      r._2.groupBy(_._2).toSeq.map { case (a, b) => a }, // Markets
      t.flatMap(_._2.groupBy(_._2).map { case (t, relation) => t }) // Tags
    )
  }.toSeq)
}

我不确定，如果我在正确的道路上。看来这几乎是我想要的。此函数将仅返回带有Records和Markets的{{1}}，而不是将它们作为可选内容。

我无法解决这个问题。似乎在任何地方都没有这种复杂查询的完整示例。任何帮助是极大的赞赏。预先感谢！

Answer 1

您的方法正确。假设您的平滑映射定义为：

case class RecordRow(id: Int)

case class TagRow(id: Int)
case class RecordTagRow(recordId: Int, tagId: Int)

case class MarketRow(id: Int)
case class RecordMarketRow(recordId: Int, marketId: Int)

class RecordTable(_tableTag: Tag)
    extends Table[RecordRow](_tableTag, "record") {
  val id = column[Int]("id", O.PrimaryKey, O.AutoInc)

  override def * = id <> ((id: Int) => RecordRow(id), RecordRow.unapply)
}

class TagTable(_tableTag: Tag) extends Table[TagRow](_tableTag, "tag") {
  val id = column[Int]("id", O.PrimaryKey, O.AutoInc)

  override def * = id <> ((id: Int) => TagRow(id), TagRow.unapply)
}

class RecordTagTable(_tableTag: Tag)
    extends Table[RecordTagRow](_tableTag, "record_tag") {
  val recordId = column[Int]("record_id")
  val tagId = column[Int]("tag_id")

  val pk = primaryKey("record_tag_pkey", (recordId, tagId))
  foreignKey("record_tag_record_fk", recordId, RecordQuery)(r => r.id)
  foreignKey("record_tag_tag_fk", tagId, TagQuery)(r => r.id)

  override def * =
    (recordId, tagId) <> (RecordTagRow.tupled, RecordTagRow.unapply)
}

class MarketTable(_tableTag: Tag)
    extends Table[MarketRow](_tableTag, "market") {
  val id = column[Int]("id", O.PrimaryKey, O.AutoInc)

  override def * = id <> ((id: Int) => MarketRow(id), MarketRow.unapply)
}

class RecordMarketTable(_tableTag: Tag)
    extends Table[RecordMarketRow](_tableTag, "record_market") {
  val recordId = column[Int]("record_id")
  val marketId = column[Int]("market_id")

  val pk = primaryKey("record_tag_pkey", (recordId, marketId))
  foreignKey("record_market_record_fk", recordId, RecordQuery)(r => r.id)
  foreignKey("record_market_market_fk", marketId, MarketQuery)(r => r.id)

  override def * =
    (recordId, marketId) <> (RecordMarketRow.tupled, RecordMarketRow.unapply)
}

val RecordQuery = new TableQuery(tag => new RecordTable(tag))
val TagQuery = new TableQuery(tag => new TagTable(tag))
val RecordTagQuery = new TableQuery(tag => new RecordTagTable(tag))
val MarketQuery = new TableQuery(tag => new MarketTable(tag))
val RecordMarketQuery = new TableQuery(tag => new RecordMarketTable(tag))

要连接具有多对多关系的表，您应该以这种方式将左联接与内部联接相结合：

val recordsQuery = RecordQuery
      .joinLeft(RecordTagQuery.join(TagQuery).on(_.tagId === _.id)).on(_.id === _._1.recordId)
      .joinLeft(RecordMarketQuery.join(MarketQuery).on(_.marketId === _.id)).on(_._1.id === _._1.recordId)

这可通过slick转换为带有PostgreSQL配置文件的以下SQL：

select
   x2."id",
   x3."id",
   x4."record_id",
   x4."tag_id",
   x3."id",
   x5."id",
   x6."record_id",
   x6."market_id",
   x5."id" 
from
   "record" x2 
   left outer join
      "record_tag" x4 
   inner join
      "tag" x3 
      on x4."tag_id" = x3."id" 
      on x2."id" = x4."record_id" 
   left outer join
      "record_market" x6 
   inner join
      "market" x5 
      on x6."market_id" = x5."id" 
      on x2."id" = x6."record_id"

最后一步是将查询结果正确映射到scala类。我是通过这种方式完成的：

db.run {
  recordsQuery.result
    .map(result => {
      result
        .groupBy(_._1._1) // RecordRow as a key
        .mapValues(values =>values.map(value => (value._1._2.map(_._2), value._2.map(_._2)))) // Seq[(Option[TagRow], Option[MarketRow])] as value
        .map(mapEntry =>(mapEntry._1, mapEntry._2.flatMap(_._1), mapEntry._2.flatMap(_._2)))  // map to Seq[(RecordRow, Seq[TagRow], Seq[MarketRow])]
        .toSeq
    })
}

这将返回Future[Seq[(RecordRow, Seq[TagRow], Seq[MarketRow])]]

Answer 2

我终于有时间再次关注这个问题。以我当前的架构和结构，我无法实现@Valerii Rusakov的答案，但是它极大地解决了问题。谢谢！

这就是我的做法：

def myFunction: Future[Seq[(RecordModel, Seq[Option[(TagsModel, Record_TagsModel)]], Seq[Option[(MarketsModel, Record_MarketModel)]], Seq[Option[(UrlsModel, Record_UrlModel)]])]] = {
val query = for {
  (((records, tags), markets), urls) <- (records filter (x => x.deleted === false && x.clientId === 1)
    joinLeft (tags join recordTags on (_.tagId === _.tagId)) on (_.recordId === _._2.recordId)
    joinLeft (markets join recordMarkets on (_.marketId === _.marketId)) on (_._1.recordId === _._2.recordId)
    joinLeft (urls join recordUrls on (_.urlId === _.urlId)) on (_._1._1.recordId === _._2.recordId))
} yield (records, tags, markets, urls)
db.run(query.result).map(_.toList.groupBy(_._1).map { // Group by records
  case (records, composedResult) =>
    (
      records,
      composedResult.groupBy(_._2).keys.toSeq, // Tags and RecordTags
      composedResult.groupBy(_._3).keys.toSeq, // Markets and RecordMarkets
      composedResult.groupBy(_._4).keys.toSeq // Urls and RecordUrls
    )
}.toSeq)
 }

请注意，我屈服于(((records, tags), markets), urls)。这使我以后可以访问那些精确的属性，从而使分组和映射变得更加容易。它仍然不是完美的，因为我必须使用表和关系表，例如TagsModel, Record_TagsModel。不过，这只是一个小问题。也许有些人知道如何解决它。当前函数返回所有records，而与tags，markets或urls无关。

光滑的嵌套外部联接与多对多表

2 个答案: