你能告诉我为什么我没有得到内心的加入我期待得到吗?
我有以下表格
case class Ability(id: UUID, can: Boolean, verb: String, subject: String, context: String)
object Abilities extends Table[Ability]("abilities"){
def id = column[UUID]("id", O.PrimaryKey)
def can = column[Boolean]("is_can")
def verb = column[String]("verb")
def subject = column[String]("subject")
def context = column[String]("context")
def * = id ~ can ~ verb ~ subject ~ context <> (Ability, Ability.unapply _)
}
case class Role(id: UUID, name : String)
object Roles extends Table[Role]("roles"){
def id = column[UUID]("id", O.PrimaryKey)
def name = column[String]("name")
def * = id ~ name <> (Role, Role.unapply _)
}
// And join table
case class AbilityRelationship(owner_id: UUID, obj_id: UUID, is_role: Boolean)
object AbilitiesMapping extends Table[AbilityRelationship]("abilities_mapping"){
def owner_id = column[UUID]("owner_id")
def obj_id = column[UUID]("obj_id")
def is_role = column[Boolean]("is_role")
def * = owner_id ~ obj_id ~ is_role <> (AbilityRelationship, AbilityRelationship.unapply _)
}
我愿意做的是为特定所有者(无论是用户还是角色)获取Ability
个对象的列表。所以我按照文档编写了以下连接查询
val some_id = role.id
val q2 = for {
a <- Abilities
rel <- AbilitiesMapping
if rel.owner_id === some_id.bind
} yield (a)
但q2.selectStatement
返回绝对错误的查询。在我的情况下,这是select x2."id", x2."is_can", x2."verb", x2."subject", x2."context" from "abilities" x2, "abilities_mapping" x3 where x3."owner_id" = ?
。
应该如何实施?
感谢。
答案 0 :(得分:8)
好吧,经过多次尝试我做到了
val innerJoin = for {
(a, rel) <- Abilities innerJoin AbilitiesMapping on (_.id === _.obj_id) if rel.owner_id === some_id.bind
} yield a
但是男人...... typesafe的文档确实对新手来说真的很弱。
答案 1 :(得分:2)
尝试类似:
val q2 = for {
a <- Abilities
rel <- AbilitiesMapping
if a.id == rel.obj_id && rel.owner_id === some_id.bind
} yield (a)
顺便说一句,你知道你可以在Table对象中注释你的外键吗?
答案 2 :(得分:1)
尝试这样做是对鲁斯兰答案的评论,但我没有足够的绝地权力:
你可以尝试一下这个desugar-ed版本吗?
val rightSide = AbilitiesMapping.filter(_.owner_id === some_id)
val innerJoin = (Abilities innerJoin (rightSide) on (
(l,r) => (l.id === r.obj_id)
).map { case (l, r) => l }