7

我正在处理地址，并且成员子前提（公寓/公寓＃）导致检索丢失。我也担心子前提是我唯一索引约束的一部分，因为它可以为空。

故障过滤器：

tableQuery.filter(c=> (c.longitude === r.longitude && c.latitude === r.latitude) ||
        (c.streetNumber === r.streetNumber && c.route === r.route && c.subpremise === r.subpremise && c.neighborhoodId === r.neighborhoodId))

成功过滤：（通过删除子前提）

tableQuery.filter(c=> (c.longitude === r.longitude && c.latitude === r.latitude) ||
            (c.streetNumber === r.streetNumber && c.route === r.route && c.neighborhoodId === r.neighborhoodId))

如果我错过了另一个促成因素，我已经包含了下面的定义，希望它会被注意到。

case class Address(id:Option[Long],streetNumber:Short,route:String,subpremise:Option[String],neighborhoodId:Fk,latitude:Option[Double],longitude:Option[Double])

class Addresses(tag: Tag) extends Table[Address](tag, "addresses") with Logging {
  def id = column[Long]("id", O.PrimaryKey, O.AutoInc)
  def streetNumber = column[Short]("street_number")
  def route = column[String]("route",O.NotNull)
  def subpremise = column[Option[String]]("subpremise")
  def neighborhoodId = column[Long]("neighborhood",O.NotNull)
  def latitude = column[Option[Double]]("latitude")
  def longitude = column[Option[Double]]("longitude")

  //Constraints
  def idx = index("idx_streetnum_route_subpremise_neighborhood", (streetNumber,route,subpremise,neighborhoodId), unique = true)
  def gps = index("gps", (latitude,longitude), unique = true)

  //Foreign Key
  def neighborhood = foreignKey("NEIGHBORHOOD_FK", neighborhoodId, Neighborhoods.tableQuery)(_.id)

  def * = (id.?,streetNumber,route,subpremise,neighborhoodId,latitude,longitude) <> (Address.tupled,Address.unapply)
}
4

2 回答 2

4

答案是使用以下检查。

( //Option Scenario both are defined
  (c.subpremise.isDefined && r.subpremise.isDefined && c.subpremise === r.subpremise) ||
  //Option Scenario both are empty
  (c.subpremise.isEmpty && r.subpremise.isEmpty)
)
于 2014-06-24T16:10:29.890 回答
0

My solution that working with Slick 3.3.x

tableQuery
  .filterIf(r.subpremise.isEmpty)(_.subpremise.isEmpty) // 1. both are empty
  .filterOpt(r.subpremise)(_.subpremise === _) // 2. both are defined and equal

Case 1 (r.subpremise is empty) corresponds the following SQL:

select * from addresses where subpremise is null

Case 2 (r.subpremise isn't empty):

select * from addresses where subpremise = ?
于 2019-09-05T10:03:07.253 回答