found : (Int, String, Option[java.lang.String])
required: (Int, String, Option[java.lang.String])
Pertinent code:
object M extends Table[(Int, String, Option[String])]("table") {
def msaid = column[Int]("msaid", O NotNull)
def name = column[String]("name", O DBType "varchar(255)")
def shape = column[Option[String]]("shape")
def * = msaid ~ name ~ shape
type T = (Int, String, Option[java.lang.String])
def apply(msa: T) = 1
def q() = db withSession { s: Session => (for (r <- M) yield M(*)).list()(s) }
^
^
...
I've also tried
type T = (Int, String, Option[String])
The ultimate goal is that I want all the selected columns to converted into an Object with named accessors, instead of being a tuple.
Scala version 2.9.2 (Java HotSpot(TM) 64-Bit Server VM, Java 1.7.0_07).
UPDATE:
Here's a Gist of the issue (slightly simplified from the above code and eliminates any String/java.lang.String "confusion" by using only Int.)