Question

我正在解码不同类型的json对象流：

type AppRow =
    | ZoneR of ZoneRow
    | CustomerR of CustomerRow

我需要返回已解码的对象：

//None work:
let fromJson x =
let fromJson x:'T =
let fromJson<'T> x =
let fromJson<'T> x:'T =
    Json.Compact.deserialize x

let decodeLog table json =
    match table with
    | "zone" ->
        fromJson json |> ZoneR
    | "customer" ->
        fromJson json |> CustomerR
    | _ -> failwith "Not ready yet"

我无法想象how replicate the effect of：

public static T deserialize<T> (string json)
{
    return JsonConvert.DeserializeObject<T> (json, Compact.Internal.Settings.settings);
}

我不想打电话给Json.Compact.deserialize因为我正在抽象编码/解码，需要在fromJson

中做一些额外的步骤

Answer 1

如果要定义一个接受字符串且具有泛型返回类型的函数，则以下两个中的任何一个都应该有效（后者允许您在调用函数时显式指定类型，前者不需要类型需要推断）。

let fromJson x : 'T = ...
let fromJson<'T> x : 'T = ...

您可以更明确地注释x，以便明确：

let fromJson (x:string) : 'T = ...

当你说你的代码不起作用时，你究竟是什么意思？以下是一个简单的愚蠢示例，只处理两个值（而不是正确解码JSON），但编译得很好并显示语法：

type AppRow =
    | ZoneR of int
    | CustomerR of string

let fromJson (x:string) : 'T =
    if x = "1" then unbox 1
    elif x = "hi" then unbox "hi"
    else failwith "wrong input"

let decodeLog table json =
    match table with
    | "zone" ->
        fromJson json |> ZoneR
    | "customer" ->
        fromJson json |> CustomerR
    | _ -> failwith "Not ready yet"  

decodeLog "zone" "1"
decodeLog "customer" "hi"

如何从函数返回F＃中的泛型值

1 个答案: