如何将键盘按键连接到操作

您需要使用常规StartApp而不是StartApp.Simple因为它提供了使用效果和任务的方法.

当按键不是空格键时,Action需要一个NoOp构造函数来允许我们的视图保持不变.

然后你需要一个将Keyboard.presses值映射到a的函数Action.这是更新的代码:

import Html exposing (div, button, text)
import Html.Events exposing (onClick)
import StartApp
import Effects exposing (Never)
import Task 
import Keyboard
import Char

app =
  StartApp.start
    { init = init
    , view = view
    , update = update
    , inputs = [ Signal.map spaceToInc Keyboard.presses ]
    }

main =
  app.html

type alias Model = Int

init =
  (0, Effects.none)

view address model =
  div []
    [ button [ onClick address Decrement ] [ text "-" ]
    , div [] [ text (toString model) ]
    , button [ onClick address Increment ] [ text "+" ]
    ]

type Action
  = Increment
  | Decrement
  | NoOp

update action model =
  case action of
    NoOp -> (model, Effects.none)
    Increment -> (model + 1, Effects.none)
    Decrement -> (model - 1, Effects.none)

spaceToInc : Int -> Action
spaceToInc keyCode =
  case (Char.fromCode keyCode) of
    ' ' -> Increment
    _ -> NoOp

port tasks : Signal (Task.Task Never ())
port tasks =
  app.tasks