SORM持久化“id”未定义

Question

我只是试着学习SORM，并玩一些我认为是简单的示例代码。使：

case class Book(var author:String, var title:String);

object Db extends Instance(
  entities = Set(Entity[Book]()),
  url = "jdbc:h2:mem:test",
  user = "",
  password = "",
  initMode = InitMode.Create
)

然后：

val b : Book with Persisted = Db.save( Book("foo","bar")  )

当我试图编译它时，我得到：

[error] /Users/rjf/IdeaProjects/PlayTest/app/controllers/Application.scala:22: type mismatch;
[error]  found   : models.Book
[error]  required: sorm.Persisted with models.Book
[error]     val b : Book with Persisted = Db.save( Book("foo","bar")  )
[error]                                                ^

如果我将Book声明更改为：

case class Book(var author:String, var title:String) extends Persisted;

然后我得到：

[error] /Users/rjf/IdeaProjects/PlayTest/app/models/Book.scala:17: class Book needs to be abstract, since:
[error] it has 2 unimplemented members.
[error] /** As seen from class Book, the missing signatures are as follows.
[error]  *  For convenience, these are usable as stub implementations.
[error]  */
[error]   def id: Long = ???
[error]   def mixoutPersisted[T]: (Long, T) = ???
[error] case class Book(var author:String, var title:String) extends Persisted;
[error]            ^

为新手的问题道歉。毫无疑问，我同时学习Scala是一个因素。

Answer 1

要解决问题，只需删除显式类型注释：

val b = Db.save( Book("foo","bar") )

别担心，您仍然可以访问与Book with Persisted相同的接口。

至于为什么会发生这种情况，这似乎是Scala和据报道的一个bug。

旁白

不要在case类声明中使用var。不变性是案例类的全部要点。正确的声明是：

case class Book(author:String, title:String)

这和

case class Book(val author:String, val title:String)