在FSharp

如何將XML序列化數組的數組這裏就是我在尋找：在FSharp

<reports> 
    <parameters> 
    <parameter name="srid" type="java.lang.Integer">16533</parameter> 
    <parameter name="pmid" type="java.lang.Integer">17018</parameter> 
    <parameter name="Start" type="java.text.SimpleDateFormat">1/1/2011 12:00:00 AM</parameter> 
    <parameter name="End" type="java.text.SimpleDateFormat">1/31/2011 12:00:00 AM</parameter> 
    </parameters> 
    <parameters> 
    <parameter name="srid" type="java.lang.Integer">16099</parameter> 
    <parameter name="pmid" type="java.lang.Integer">17018</parameter> 
    <parameter name="Start" type="java.text.SimpleDateFormat">1/1/2011 12:00:00 AM</parameter> 
    <parameter name="End" type="java.text.SimpleDateFormat">1/31/2011 12:00:00 AM</parameter> 
    </parameters> 
</reports>

而是，這裏就是我得到：

<reports> 
    <parameters> 
     <parameters> 
     <parameter name="srid" type="java.lang.Integer">16533</parameter> 
     <parameter name="pmid" type="java.lang.Integer">17018</parameter> 
     <parameter name="Start" type="java.text.SimpleDateFormat">1/1/2011 12:00:00 AM</parameter> 
     <parameter name="End" type="java.text.SimpleDateFormat">1/31/2011 12:00:00 AM</parameter> 
     </parameters> 
    </parameters> 
    <parameters> 
     <parameters> 
     <parameter name="srid" type="java.lang.Integer">16677</parameter> 
     <parameter name="pmid" type="java.lang.Integer">17018</parameter> 
     <parameter name="Start" type="java.text.SimpleDateFormat">1/1/2011 12:00:00 AM</parameter> 
     <parameter name="End" type="java.text.SimpleDateFormat">1/31/2011 12:00:00 AM</parameter> 
     </parameters> 
    </parameters> 
</reports>

看來我額外的<parameters>標籤。

這裏是我的整個模型：

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 parameters(parameters: parameter array) = 
    let mutable paramArray = parameters 

    public new() = 
     new parameters(Array.empty) 

    [<XmlArray "parameters">] 
    member this.ParamArray with get() = paramArray and set v = paramArray <- v 

[<XmlRoot("reports")>] 
type reports(ps:parameters array) = 
    let mutable parms = ps 

    public new() = 
     new reports(Array.empty) 

    [<XmlElement("parameters")>] 
    member this.Ps with get() = parms and set v = parms <- v

來源

2011-02-08 Ramy

這是否對您有幫助？ http://stackoverflow.com/questions/2928108/how-to-serialize-this-xml-in-net-array – Brian 2011-02-08 20:25:08

我不相信它。檢查我的更新。我用第二組參數進行了澄清。 – Ramy 2011-02-08 20:32:06

好了，我簡化你的類型有點看清潔：

type parameter(paramName) = 
    let mutable pName = paramName 

    public new() = 
    new parameter("") 

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


type parameters(parameters: parameter array) = 
    let mutable paramArray = parameters 

    public new() = 
    new parameters(Array.empty) 

    [<XmlElement "parameter">] 
    member this.ParamArray with get() = paramArray and set v = paramArray <- v 

[<XmlRoot("reports")>] 
type reports(ps:parameters array) = 
    let mutable parms = ps 

    public new() = 
    new reports(Array.empty) 

    [<XmlElement("parameters")>] 
    member this.Ps with get() = parms and set v = parms <- v

，然後串行化這些：

let params1 = parameters ([|parameter("a"); parameter("b")|]) 
    let params2 = parameters ([|parameter("c"); parameter("d")|]) 

    let repos = reports ([|params1; params2|]) 

    use writer = System.Xml.XmlWriter.Create @"C:\temp\foo1.xml" 
    let xs = System.Xml.Serialization.XmlSerializer typeof<reports> 
    xs.Serialize (writer, repos)

產生：

<reports xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> 
    <parameters> 
     <parameter name="a" /> 
     <parameter name="b" /> 
    </parameters> 
    <parameters> 
     <parameter name="c" /> 
     <parameter name="d" /> 
    </parameters> 
</reports>

hth，alex

來源

2011-02-08 20:54:39 Alex

回答

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