CODEWITHSUNDEEP
scalasqueryl
xyzlast
xyzlast

Reputation: 690

how to implement @MappedSuperClass in squeryl?

I have 12 tables in the database. and all tables has same columns and same relations.

So I had writen the code in JPA. 

@MappedSuperclass
public class BaseAddress {
... properties and relations (@ManyToOne / @OneToMany)
}

@Entity
public class Address1 extends BaseAddress {
  // has no code. properties and relations are inherited.
}

@Entity 
public class Address2 extends BaseAddress {
  // has no code. properties and relations are inherited.    
}

but, I don't know how implemented it in squeryl schema definition. How implement that relation inheritance in squeryl ?

Upvotes: 0

Views: 110

Answers (1)

Dave Whittaker
Dave Whittaker

Reputation: 3102

Squeryl will reflect on all of the available fields. It doesn't make a distinction on whether they are inherited or not. Something like:

class BaseAddress(id: Long, address: String)

class Address1(id: Long, address: String) extends BaseAddress(id, address)

class Address2(id: Long, address: String) extends BaseAddress(id, address)

object MySchema extends Schema {

  val address1 = table[Address1]

  val address2 = table[Address2]

}

Should work similarly to the Hibernate code you posted. For the relationships, with Squeryl they are defined in the Schema object rather than within the model class. You can add a helper method if you're defining the same relation more than once:

def addressTableWithUserRelation[T <: BaseAddress]: (Table[T], OneToManyRelation[User, T]) = {
  val table = Table[T]
  val relation = oneToManyRelation(users, table)((u,t) => u.id === t.userId)
  (table, relation)
}

val (address1, address1ToUser) = addressTableWithUserRelation[Address1]

Upvotes: 2

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