CODEWITHSUNDEEP
scalaplayframework-2.0slick
user504909
user504909

Reputation: 9549

why does the slick can not store Option[String]?

I use slick 2.0 rc1 in my playframework 2.2 project my talbe code is:

case class Resource(id: Option[Long] = None, owner: UserId, types: String)

// The static object that does the actual work - note the names of tables and fields in H2 are case sensitive and must be all caps 
object Resources extends Table[Resource]( "RESOURCE") {
  def id = column[Long]("ID", O.PrimaryKey, O.AutoInc)
  def owner = column[UserId]("Owner")
  def types = column[String]("Type")
  def withuser = foreignKey("User_FK", owner, Users)(_.id)

 // Every table needs a * projection with the same type as the table's type parameter
 def * = id.? ~ owner ~ types <> (Resource, Resource.unapply _)
}

the output error is:

[info] Compiling 16 Scala sources and 2 Java sources to C:\assigment\slick-advan
ced\target\scala-2.10\classes...
[error] C:\assigment\slick-advanced\app\models\Resource.scala:12: overloaded met
hod constructor Table with alternatives:
[error]   (_tableTag: scala.slick.lifted.Tag,_tableName: String)play.api.db.slic
k.Config.driver.Table[models.Resource] <and>
[error]   (_tableTag: scala.slick.lifted.Tag,_schemaName: Option[String],_tableN
ame: String)play.api.db.slick.Config.driver.Table[models.Resource]
[error]  cannot be applied to (String)
[error] object Resources extends Table[Resource]( "RESOURCE") {
[error]                  ^

I know on document it like:

 object Resources(tag: Tag) extends TableTable[(Long, UserId, String)](tag, "RESOURCE") {

but after I change it, it still have error:

[error] C:\assigment\slick-advanced\app\models\Resource.scala:12: traits or obje
cts may not have parameters
[error] object Resources(tag: Tag) extends TableTable[(Long, UserId, String)](ta
g, "RESOURCE") {
[error]                 ^
[error] one error found

Upvotes: 1

Views: 281

Answers (1)

Vidya
Vidya

Reputation: 30310

Check out the Scaladoc for Table for the constructors:

new Table(_tableTag: Tag, _tableName: String)
new Table(_tableTag: Tag, _schemaName: Option[String], _tableName: String)

The compiler is telling you that you have those choices but are using neither. You construct your instance of Table called Resources with just a String. As you can see, that isn't an option. No pun intended.

The documentation shows that your declaration should look more like this:

class Resources(tag: Tag) extends Table[(Long, UserId, String)](tag, "RESOURCE") {
  ...
}
val Resources = TableQuery[Resources]

Upvotes: 1

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