Question

我需要为一个序列的每个元素调用一个函数，目前我已经尝试过了 Seq.iter和Seq.map但它们分别返回单位和＆＃39; a - ＆gt;＆＃39; c而不是我需要的Json。

我试过了

Seq.iter (fun _ (a,b,c,d) -> iterateThroughMySequnce a b c d ()) sequence
Seq.fold (fun _ (a,b,c,d) -> iterateThroughMySequnce a b c d ()) sequence

但我没有得到Json的预期返回类型。代码需要在下面的注释中说明＆＃34;像这样的东西＆＃34;

任何人都可以帮助谢谢

open Newtonsoft.Json.Linq

type Json =
    | JObj of Json seq
    | JProp of string * Json
    | JArr of Json seq
    | JVal of obj

let (!!) (o: obj) = JVal o

let rec toJson = function
    | JVal v -> new JValue(v) :> JToken
    | JProp(name, (JProp(_) as v)) -> new JProperty(name, new JObject(toJson v)) :> JToken
    | JProp(name, v) -> new JProperty(name, toJson v) :> JToken
    | JArr items -> new JArray(items |> Seq.map toJson) :> JToken
    | JObj props -> new JObject(props |> Seq.map toJson) :> JToken

let sequence = seq { yield "USD", 12.36M, 156.32M, 18.23M
                     yield "JPY", 13.36M, 564.32M, 17.23M 
                     yield "GBP", 14.36M, 516.32M, 120.23M }

let iterateThroughMySequnce a b c d () =
    JObj [JProp("CurrencyCode", !! a);
          JProp("TotalPerCurrencyBeforeExchange", !! b); 
          JProp("ExchangeRate", !! c);
          JProp("TotalPerCurrencyAfterExchange", !! d)];

let k =
    JObj [
        JProp("InvoiceNumber", !! "13456789");
        JProp("InvoiceDate", !! "21/12/2015");
        JProp("InvoiceCurrency", !! "USD");
        JProp("InvoiceProfitMargin", !! 2.3);
        JProp("InvoicePaymentCurrencyToEuroExchangeRate", !! 0.8658745M);
        JProp("InvoicePeroid", !! "01/01/2015 00:00:00 - 01/02/2015 23:59:59");
        JProp(
            "Transaction", 
                JArr [
                    //Something like this
                    Seq.iter (fun (a,b,c,d) -> iterateThroughMySequnce a b c d ()) sequence
                ])
        JProp("TransactionForPeroid", !! 254584.00M);
        JProp("InvoicingAmountWithProfitMarginApplied", !! 8452.01M);
        JProp("InvoicingAmountWithProfitMarginAppliedInEuro", !! 7851.28);
    ]

let json = toJson k

Answer 1

您需要Seq.map，它将输入序列转换为输出序列（并使用指定的函数将每个元素转换为新值）。您的代码几乎是正确的，但调用不应该包含在另一个列表中：

JProp(
    "Transaction", 
        JArr (Seq.map (fun (a,b,c,d) -> iterateThroughMySequnce a b c d ()) sequence)
)

如果你改变你的iterateThroughMySequence函数来接受一个元组（并且它的名称应该不同，因为它不是迭代！），你可以做得更好。

let formatItemAsJson (a,b,c,d) =
  JObj [JProp("CurrencyCode", !! a);
        JProp("TotalPerCurrencyBeforeExchange", !! b); 
        JProp("ExchangeRate", !! c);
        JProp("TotalPerCurrencyAfterExchange", !! d)];

// Later in the main part of code
JProp("Transaction", JArr (Seq.map iterateThroughMySequnce sequence))

除此之外，F＃数据库还带有JsonValue类型（请参阅the API reference），它实现了您在此处所做的一些事情 - 它可以让您构建＆amp;格式化JSON（并解析它）。

F＃遍历序列并为序列的每个元素调用一个函数

1 个答案: