Scala 光滑模式leftJoin生成笛卡尔积
我有两张桌子:Scala 光滑模式leftJoin生成笛卡尔积,scala,playframework,slick,scalaquery,Scala,Playframework,Slick,Scalaquery,我有两张桌子: select * from "DEPARTMENTS"; ID | NAME | MANAGER_ID ----+-------+------------ 1 | FOO | 1 3 | XXX | 2 4 | dept1 | (3 rows) select * from "EMPLOYEES"; NAME | LAST | EMAIL | PHONE | SKY
select * from "DEPARTMENTS";
ID | NAME | MANAGER_ID
----+-------+------------
1 | FOO | 1
3 | XXX | 2
4 | dept1 |
(3 rows)
select * from "EMPLOYEES";
NAME | LAST | EMAIL | PHONE | SKYPE | DEPT_ID | ID
------+------+------------------+-------+-------+---------+----
AAA | xxxx | abcdef@gmail.com | | | | 1
BBB | yyyy | | | | | 2
(2 rows)
如果经理总共有3行,我想得到所有有经理名字的部门
但是,在我的模式中,当我这样做时:
import scala.slick.driver.PostgresDriver.simple._
import play.api.Play.current
case class DepartmentManager(id:Int,name:String,manager:String="")
case class Department(id:Option[Int],name:String,managerId:Option[Int])
class Departments (tag: Tag) extends Table[Department](tag, "DEPARTMENTS") {
val employees = TableQuery[Employees]
def id = column[Int]("ID", O.PrimaryKey, O.AutoInc)
def name = column[String]("NAME", O.NotNull)
def managerId = column[Int]("MANAGER_ID", O.Nullable)
def manager = foreignKey("EMP_FK",managerId,employees)(_.id)
def * = (id.?,name,managerId.?) <> (Department.tupled, Department.unapply)
}
object Departments{
val db = play.api.db.slick.DB
val departments = TableQuery[Departments]
//this is the problematic query
def allWithMngr = db.withSession { implicit session =>
val employees = TableQuery[Employees]
val dm = (departments leftJoin employees ).list
dm.map(x =>DepartmentManager(x._1.id.getOrElse(0),x._1.name, x._2.name + " " + x._2.last))
}
}
结果出现错误:
[SlickException:读取结果集列路径的NULL值NULL
s2._4]
这是我的员工模型:
case class Employee(name: String,last: String,email:Option[String]=None,phone:Option[String]=None,skype:Option[String]=None,department: Option[Int] = None, id: Option[Int] = None)
class Employees (tag: Tag) extends Table[Employee](tag, "EMPLOYEES") {
def id = column[Int]("ID", O.PrimaryKey, O.AutoInc)
def name = column[String]("NAME", O.NotNull)
def last = column[String]("LAST", O.NotNull)
def email = column[String]("EMAIL", O.Nullable)
def phone = column[String]("PHONE", O.Nullable)
def skype = column[String]("SKYPE", O.Nullable)
def deptId = column[Int]("DEPT_ID", O.Nullable)
def dept = foreignKey("EMP_FK",deptId,departments)(_.id)
def * = (name,last,email.?,phone.?,skype.?,deptId.?, id.?) <> (Employee.tupled, Employee.unapply)
val departments = TableQuery[Departments]
}
这将导致:
List (DepartmentManager(1,"FOO","AAA"),DepartmentManager(3,"XXX","BBB"),DepartmentManager(4,"FOO",""))
你在这里遇到的是Slick 2处理连接中空值的笨拙方式。在Slick 3中情况更好,但我们可以在Slick 2中加入工作 您的查询
(departments leftJoin employees ).list
…将是交叉连接,因为您没有连接
使用联接时遇到的错误是因为要选择的列之一为NULL。使用Slick 2,您必须明确选择这些列,并使用来提升它们。?选择一个选项:
val query =
departments.leftJoin(employees).on(_.managerId === _.id )
.map { case (d, e) => (d.id, d.name, e.name.?) }
注意e.name.?-因为该部门可能没有e员工
此查询与测试数据的结果为:
(1, FOO, Some(AAA))
(2, XXX, Some(BBB))
(3, dept1, None)
…我想这正是您想要的。您能以表格形式给出您想要的输出示例吗?离题:如果数据库设计尚未解决,我建议使用关系表而不是可为空的外键。这样可以避免一些边缘情况,并简化查询。@Jus12我现在做了,谢谢你的评论。除了SQL之外,表格式将更有帮助。我们需要的是实际的查询和返回的数据。@Jus12它实际上在查询列表departmentmanager下面……谢谢,这正是我要找的。现在清楚多了
val query =
departments.leftJoin(employees).on(_.managerId === _.id )
.map { case (d, e) => (d.id, d.name, e.name.?) }
(1, FOO, Some(AAA))
(2, XXX, Some(BBB))
(3, dept1, None)