scala – 如何在Slick中正确比较Options成员？

发布时间：2020-12-16 18:35:56 所属栏目：安全来源：网络整理

导读：我正在使用地址做事,而成员子预设(公寓/公寓#)导致检索错过.我也担心subpremise是我唯一索引约束的一部分,因为它可以为null. 故障过滤器： tableQuery.filter(c= (c.longitude === r.longitude c.latitude === r.latitude) || (c.streetNumber === r.streetN

我正在使用地址做事,而成员子预设(公寓/公寓#)导致检索错过.我也担心subpremise是我唯一索引约束的一部分,因为它可以为null.

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removung subpremise)

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

我已经把这些定义包括在下面了.如果还有其他因素我错过了,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,latitude,longitude) <> (Address.tupled,Address.unapply)
}

解决方法

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)
)

（编辑：李大同）

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