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如何从列表和(a->Bool)中构建SelectList?

elm

Natim · 技术社区 · 6 年前

我正在使用 elm-spa-example 模特和我有一个 Session 包含 List Recipe 以及包含 recipeID

我想建立一个 SelectList 这将选择列表中包含RecipeID的配方。

理想情况下,我会使用以下内容:

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

我的情况我会:

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

3 回复 | 直到 6 年前

Natim 6 年前

我是这样做的:

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

我还补充道:

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

stevensonmt 6 年前

我把这个快速的ellie放在一起,它说明了我认为实现你想要的目标所需的步骤。它当然不是优化的,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"

JosephStevens 6 年前