f# - 如何像组件框架一样将逻辑封装在子级中?
问题描述
我试图了解如何使用 WebSharper 的 F# Bolero 中的 Elmish 架构创建可重用组件(例如,可重用的验证表单输入)。从我看到的所有示例中,顶级父级必须处理所有消息/更新和逻辑,而子级仅用于视图。我想知道是否有办法解决这个问题,是否让孩子处理自己的状态+消息,并将某些消息传播给父母(我在下面的代码中尝试过),或者是否有另一种设计来处理这个问题。
在我的具体情况下,我正在尝试为验证两个字段都不为空的用户名创建表单输入组件。我不喜欢让父处理更新各个字段 FirstName 和 LastName 的想法,它应该只关心获取 Submit 消息。如果您多次使用孩子,处理孩子产生的每条消息都会导致大量样板文件
注意:我提供的代码无法编译,因为我正在努力理解如何实现我的预期设计
open Elmish
open Bolero
open Bolero.Html
module NameInput =
type Model = { FirstName : string; LastName : string }
type Message =
| ChangeFirstName of string
| ChangeLastName of string
| Submit of Model
let update model msg =
match msg with
| ChangeFirstName s ->
{ model with FirstName = s}, Cmd.none
| ChangeLastName s ->
{ model with LastName = s}, Cmd.none
| Submit m ->
m, Cmd.ofMsg (Submit m)
type Component() =
inherit ElmishComponent<Message, Model>()
let invalidField s = s <> ""
override this.View model dispatch =
let fnClass = if (invalidField model.FirstName) then "invalid" else "valid"
let lnClass = if (invalidField model.LastName) then "invalid" else "valid"
div [] [
label [] [ text "First Name: " ]
input [
attr.``class`` fnClass
on.change (fun e -> update model (ChangeFirstName (unbox e.Value)))
]
label [] [ text "Last Name: " ]
input [
attr.``class`` lnClass
on.change (fun e -> update model (ChangeLastName (unbox e.Value)))
]
button [ on.click (fun _ -> update model (Submit model)) ] [ text "Submit" ]
]
type Message =
| NameSubmitted of NameInput.Message.Submit
type Model = { UserName : NameInput.Model }
let initModel = { UserName = { FirstName = ""; LastName = "" } }
let update msg model =
match msg with
| NameSubmitted name ->
// Greet the user
{ model with UserName = name }, Cmd.none
let view model dispatch =
concat [
ecomp<NameInput.Component,_,_>
model.Username dispatch
]
type MyApp() =
inherit ProgramComponent<Model, Message>()
override this.Program =
Program.mkProgram (fun _ -> initModel, Cmd.none) update view
解决方案
感谢@rmunn 和@hvester 的参考,它帮助我更好地了解 Elmish 并能够提出解决方案。作为可能偶然发现此问题的任何其他人的参考,这里是解决方案。InternalMessage 不需要私有,它只是将这些情况从主程序的更新功能中隐藏起来,这样人们就可以很容易地看到他们需要处理哪些消息。如果它是公开的,如果您尝试匹配 InternalMessage 案例而不首先将 Message 解包到 InternalMessage 中,编译器将给出错误(因此程序员仍然很容易知道哪些消息是内部的)
module NameInput =
type Model = { FirstName : string; LastName : string }
type private InternalMessage =
| ChangeFirstName of string
| ChangeLastName of string
type Message =
| Internal of InternalMessage
| Submit of Model
let update msg model =
match msg with
| ChangeFirstName s ->
{ model with FirstName = s }
| ChangeLastName s ->
{ model with LastName = s }
type Component() =
inherit ElmishComponent<Model, Message>()
let invalidField s = s <> ""
override this.View model dispatch =
let fnClass = if (invalidField model.FirstName) then "invalid" else "valid"
let lnClass = if (invalidField model.LastName) then "invalid" else "valid"
div [] [
label [] [ text "First Name: " ]
input [
attr.``class`` fnClass
on.change (fun e -> dispatch << Internal << ChangeFirstName <| unbox e.Value)
]
label [] [ text "Last Name: " ]
input [
attr.``class`` lnClass
on.change (fun e -> dispatch << Internal << ChangeLastName <| unbox e.Value)
]
button [ on.click (fun _ -> dispatch <| Submit model) ] [ text "Submit" ]
]
type Model = { Name : NameInput.Model }
let initModel = { Name = { FirstName = ""; LastName = "" } }
type Message =
| NameInput of NameInput.Message
let update message model =
match message with
| NameInput ni ->
match ni with
| NameInput.Internal i ->
{ model with Name = model.Name |> NameInput.update i}
| NameInput.Submit n ->
{ model with Name = n }
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