语法：返回F＃​​中已区分联合的参数的函数？ 功能参数的模式匹配

问题描述

我显然是F＃的新手。来自C＃，我在使用歧视工会方面遇到困难。

假设在F＃中具有以下定义：

type Details =
            | ContactDetails of ContactDetail * id: Guid
            | Internet       of Internet  * id: Guid    
            | PhoneNumbers   of PhoneNumber  * id: Guid 
            | Addresses      of Address     * id: Guid  
    
        let contactDetail  : ContactDetail = {Name="Contact Detail"; Content="Content for Contact Detail"; Text="here is the contact detail text" }    
        let internet       : Internet = {Name="Internet";       Content="Content for Internet";       Text="here is the internet text" }
        let phoneNumber    : PhoneNumber =  {Name="Phone Number";   Content="Content for phone number";   Text="here is the phone number text" }
        let address        : Address = {Name="Address";        Content="Content for Address";        Text="here is the Address text" }
       
        let details   = [ContactDetails (contactDetail,Guid.NewGuid())
                         Internet       (internet,Guid.NewGuid())
                         PhoneNumbers   (phoneNumber,Guid.NewGuid())
                         Addresses      (address,Guid.NewGuid())
                         ]


type Model = {
      Details: Details list
    }

我该如何编写包含Model并返回列表中每个项目的ID的函数？

即，例如：

乐趣细节-> detail.id

类型“ Details”未定义字段，构造函数或成员“ id”

TIA

编辑＃1：

新型号为：

type Model = { 详细信息：DetailsWithId列表 }

解决方法

“有区别的联盟”中的“有区别的”意味着每个字段定义都被认为是彼此不同的，这就是为什么即使您有一个公共字段也无法直接访问它。

我通常要做的最简单的事情是创建一个扩展函数，以安全地提取该字段。

type Details =
            | ContactDetails of ContactDetail * id: Guid
            | Internet       of Internet  * id: Guid    
            | PhoneNumbers   of PhoneNumber  * id: Guid 
            | Addresses      of Address     * id: Guid  

    member this.id = 
        match this with
        | ContactDetail(_,id)
        | Internet(_,id)
        | PhoneNumber(_,id)
        | Address(_,id) -> id

另一种选择是以不同的方式构造数据：

type Details =
     | ContactDetails of ContactDetail
     | Internet       of Internet
     | PhoneNumbers   of PhoneNumber
     | Addresses      of Address

// Each instance will hold one of ContactDetails,Internet,Phonenumber,and
// each has a common property of ID (the Guid). This way,you simplify your model
type DetailsWithId = DetailsWithId of Details * Guid

另一句话：您对DU的每个成员使用复数形式，但该定义仅允许一项。