您如何迭代列表（也许一个）

Question

我有以下graphQL结果：

[仅{详细信息=仅“ 33秒引擎故障和车辆丢失”，launch_year =仅“ 2006”，链接=仅{article_link =仅“ https://www.space.com/2196-spacex-inaugural-falcon -1-rocket-lost-launch.html “}，mission_name = Just” FalconSat“}]

基于以下类型：

type alias Launch =
    { mission_name : Maybe String
    , details : Maybe String
    , launch_year : Maybe String
    , links : Maybe LaunchLinks
    }


type alias Launches =
    Maybe (List (Maybe Launch))


type alias LaunchLinks =
    { article_link : Maybe String
    }

我想通过List.map并将结果显示在无序列表中。我从这个开始：

renderLaunch : Launches -> Html Msg
renderLaunch launches =
    div [] <|
        case launches of
            Nothing ->
                [ text "Nothing here" ]

            Just launch ->
                launch
                    |> List.map (\x -> x)
                    |> ul []

但我不断收到此错误：

此函数无法处理通过（|>）管道发送的参数：

141 | 发射142 | |> List.map（\ x-> x）143 | |> ul [] ^^^^^参数为：

List (Maybe Launch)

但是（|>）将它传递给以下函数：

List (Html msg)

Answer 1

问题是这种Just launch情况需要导致a，List (Html msg)但是代码会导致返回不同的类型。

使用时List.map (\x -> x)，它本质上是无操作的。您正在遍历a List (Maybe Launch)并返回相同的东西。我建议创建另一个带有Maybe Launch值的函数并将其用作映射函数。例如：

displayLaunch : Maybe Launch -> Html Msg
displayLaunch launch =
    case launch of
        Nothing -> text "No launch"
        Just l -> text (Debug.toString l)

现在，您可以将其插入到映射功能中：

Just launch ->
    launch
        |> List.map displayLaunch
        |> ul []

但是，哎呀！现在，您将收到一个新错误，指示：

The 2nd branch is:

    Html Msg

But all the previous branches result in:

    List (Html msg)

这里的问题是我们现在ul从Just launch分支返回a ，并且我们需要返回html列表。您可以List.singleton用来创建仅包含一项的列表：

The 2nd branch is:

    Html Msg

But all the previous branches result in:

    List (Html msg)