.net 使用FSharp的XML序列化

我正在尝试创建一些XML，如下所示：

<parameter name="srid" type="java.lang.Integer">24729</parameter>

24729

请注意，标记名为parameter，除了实际值之外，它还有两个属性、一个名称和一个类型

以下是我到目前为止的情况：

type parameter(paramName, javaType, paramValue) =
    let mutable pName = paramName
    let mutable pType = javaType
    let mutable pValue = paramValue

    public new() = 
        new parameter("","","")

    [<XmlAttributeAttribute("name")>]    
    member this.PName with get() = pName and set v = pName <- v

    [<XmlAttributeAttribute("type")>]
    member this.PType with get() = pType and set v = pType <- v

type参数（paramName、javaType、paramValue）=
让可变pName=paramName
让可变pType=javaType
设可变pValue=paramValue
公共新
新参数（“，”，“”）
[]    
使用get（）=PName成员this.PName并设置v=PNameok，因此需要

[<XmlText>]

[]

要序列化节点值的属性的属性

所以，在你的情况下，你会有

type parameter(paramName, javaType, paramValue) =
    let mutable pName = paramName
    let mutable pType = javaType
    let mutable pValue = paramValue

    public new() = new parameter("","","")

    [<XmlAttributeAttribute("name")>]    
    member this.PName with get() = pName and set v = pName <- v

    [<XmlAttributeAttribute("type")>]
    member this.PType with get() = pType and set v = pType <- v

    [<XmlText>]
    member this.PValue with get() = pValue and set v = pValue <- v

type参数（paramName、javaType、paramValue）=
让可变pName=paramName
让可变pType=javaType
设可变pValue=paramValue
public new（）=新参数（“”，“”，“”）
[]    
使用get（）=PName和set v=PName成员this.PName非常有帮助，谢谢Alex。如果你能帮忙，我还有一个问题：
type parameter(paramName, javaType, paramValue) =
    let mutable pName = paramName
    let mutable pType = javaType
    let mutable pValue = paramValue

    public new() = new parameter("","","")

    [<XmlAttributeAttribute("name")>]    
    member this.PName with get() = pName and set v = pName <- v

    [<XmlAttributeAttribute("type")>]
    member this.PType with get() = pType and set v = pType <- v

    [<XmlText>]
    member this.PValue with get() = pValue and set v = pValue <- v