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Storing case object with squeryl

Tags:

scala

squeryl

How do I store user case objects with squeryl? I have an Account object with a permission field of type Permission (defined as a sealed trait). I also have 2 case objects (Administrator and NormalUser) extending from Permission. How can I persist the Account class using Squeryl. Example code below:

sealed trait Permission
case object Administrator extends Permission
case object NormalUser extends Permission

case class Account(
        id: Long, 
        email: String,
        permission: Permission
) extends KeyedEntity[Long]
like image 686
jorgenfb Avatar asked May 21 '26 20:05

jorgenfb


1 Answers

Expanding on my comment, if you use a custom type to retrieve the permission type, such that it persists to the database as an integer (in the example below 1 and 0), you can override the unapply method to lookup the case object and pattern matching should work fine. I imagine something like the following should work:

class Permission(identifier:Int) extends org.squeryl.customtypes.IntField(identifier) {
  self: CustomType[Int] =>

  private lazy val permissions = 
    List(Administrator, NormalUser).
      map(p => p.value -> p).
      toMap

  def unapply = permissions.get(value)
}

case object Administrator extends Permission(1)
case object NormalUser extends Permission(0)

Then you should be able to store the permission directly in your code, using your entity definition:

case class Account(
        id: Long, 
        email: String,
        permission: Permission
) extends KeyedEntity[Long]

you can set the permission field directly as Administrator or NormalUser and you should also be able to pattern match like:

account.permission match {
  case Administrator => ..
  case NormalUser => ..
}
like image 122
jcern Avatar answered May 23 '26 10:05

jcern