found : (Int, String, Option[java.lang.String]) required: (Int, String, Option[java.lang.String])
Relevant 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 also tried
type T = (Int, String, Option[String])
The ultimate goal is that I want all the selected columns to be converted to an object with named accessor, 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 the Gist problem (slightly simplified from the code above and eliminates any confusion String / java.lang.String "using only Int.)
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