Question

我使用的是elm-spa-example模型，我的Session包含List Recipe以及包含recipeID的模型

我想构建一个SelectList来选择已在列表中获得RecipeID的食谱。

理想情况下，我会使用类似的东西：

SelectList.selectFromList : (a -> Bool) -> List a -> SelectList a

我的情况我会这样做：

SelectList.selectFromList (\recipe -> recipe.id == recipeID) session.recipes

Answer 1

我做了类似的事情：

selectFromList : (a -> Bool) -> List a -> Maybe (SelectList a)
selectFromList isSelectable list = 
    case list of
        first :: rest ->
            SelectList.fromLists [] first rest
                |> SelectList.select isSelectable
                |> Just
        [] ->
            Nothing

我还补充说：

prev : SelectList a -> Maybe a
prev list =
    SelectList.before list
        |> List.reverse
        |> List.head


next : SelectList a -> Maybe a
next list =
    SelectList.after list
        |> List.head

Answer 2

我把这个快速的ellie放在一起，说明了我认为实现你想要的步骤所需的步骤。它肯定没有优化，甚至不是惯用的。

https://ellie-app.com/4TJVgSCwXa1/0

firstPartialList = takeWhile condition myList
selected = Maybe.withDefault "" (getAt (length firstPartialList) myList)
secondPartialList = drop ((length firstPartialList) + 1) myList

mySelectList = SelectList.fromLists firstPartialList selected secondPartialList

condition = (\item -> item /= otherItem)


myList = ["a", "b", "c", "d"]

otherItem = "b"

Answer 3

SelectList没有公开selectFromList你确定链接是对的吗？

如何从列表和（a - ＆gt; Bool）构建一个SelectList？

3 个答案: