CODEWITHSUNDEEP
scalascala-swing
David Matuszek
David Matuszek

Reputation: 1263

How to create an Option dialog in Scala

I've been trying to create an Option dialog that isn't limited to two or three choices (Option.YesNo or Option.YesNoCancel), but I've been unable to find a way to use anything but these built-in options. Specifically, the following refuses to accept anything that I can put for optionType:

    object Choices extends Enumeration {
      type Choice = Value
      val red, yellow, green, blue = Value
    }
        val options = List("Red", "Yellow", "Green", "Blue")
        label.text = showOptions(null,
                    "What is your favorite color?",
                    "Color preference",
                    optionType = Choices.Value,
                    entries = options, 
                    initial = 2) match {
          case Choice.red => "Red"
          case Choice.yellow => "Yellow"
          case Choice.green => "Green"
          case Choice.blue => "Blue"
          case _ => "Some other color"
        }

Upvotes: 2

Views: 810

Answers (2)

0__
0__

Reputation: 67330

Yes, it's one of the many design bugs in Scala-Swing. You can write your own showOptions method:

import swing._
import Swing._
import javax.swing.{UIManager, JOptionPane, Icon}

def showOptions[A <: Enumeration](
                parent: Component = null, 
                message: Any, 
                title: String = UIManager.getString("OptionPane.titleText"),  
                messageType: Dialog.Message.Value = Dialog.Message.Question, 
                icon: Icon = EmptyIcon, 
                entries: A,
                initial: A#Value): Option[A#Value] = {
  val r = JOptionPane.showOptionDialog(
                  if (parent == null) null else parent.peer,  message, title, 0, 
                  messageType.id, Swing.wrapIcon(icon),  
                  entries.values.toArray[AnyRef], initial)
  if (r < 0) None else Some(entries(r))
}

val res = showOptions(message = "Color", entries = Choices, initial = Choices.green)

If you want to pass in strings instead, change to entries: Seq[Any], initial: Int, use entries(initial) in the call, and just return r of Int.

Upvotes: 1

Robin Green
Robin Green

Reputation: 33093

optionType isn't for passing in a Scala type, and actually I'm not sure what its purpose is in this API. But you can just set optionType = Options.YesNoCancel, and that should work, I think.

Upvotes: 0

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