F# Yaml 类型提供程序

Question

我尝试在我的配置文件中使用 Yaml 地图集合：

Companies:
  - code: 11
    name: A
    country: FR
    functionalCurrency: EUR
  - code: 12
    name: B
    country: GB
    functionalCurrency: GBP

但是，当尝试使用类型提供程序读取它时，它只会找到列表的第一个结果。

与：

  open FSharp.Configuration
  type CompaniesConfig = YamlConfig<"Config.yaml">  
  let config = CompaniesConfig()

输出是：

  val config : CompaniesConfig =
    Companies:
    - code: 11
      name: A
      country: FR
      functionalCurrency: EUR

尝试在线解析代码有效，因此我想知道这是库限制还是...？

感谢您的帮助

Answer 1

您需要实际加载文件，如果您想直接使用它，不仅要获取架构：config.Load(yamlFile)。这在文档中应该更明确。我使用了 link 中的示例文件。

#if INTERACTIVE
#r @"..\packages\FSharp.Configuration.0.6.1\lib\net40\FSharp.Configuration.dll"
#endif

open FSharp.Configuration
open System.IO

/// https://github.com/fsprojects/FSharp.Configuration/blob/master/tests/FSharp.Configuration.Tests/Lists.yaml

[<Literal>]
let yamlFile = __SOURCE_DIRECTORY__ + "..\Lists.yaml"

File.Exists yamlFile

type TestConfig = YamlConfig<yamlFile>
let config = TestConfig()

config.Load(yamlFile)
config.items.Count
config.items

我得到了两个项目：

> 
val it : int = 2
> 
val it : System.Collections.Generic.IList<TestConfig.items_Item_Type> =
  seq
    [FSharp.Configuration.TestConfig+items_Item_Type
       {descrip = "Water Bucket (Filled)";
        part_no = "A4786";
        price = 147;
        quantity = 4;};
     FSharp.Configuration.TestConfig+items_Item_Type
       {descrip = "High Heeled "Ruby" Slippers";
        part_no = "E1628";
        price = 10027;
        quantity = 1;}]
>

F# Yaml 类型提供程序

F# Yaml type provider

f#

yaml

fsharp.data.typeproviders