如何正确反序列化XML属性和数组？

Question

我正在做一个C＃项目，我有一个用XML编码的对象。 一个示例实例为：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<Entity Type="StartRunTestSetResponse">
    <Fields>
        <Field Name="SuccessStaus">
            <Value>2</Value>
        </Field>
        <Field Name="info">
            <Value></Value>
        </Field>
    </Fields>
</Entity>

我需要属性信息，因为它是对象具有的键-值对的必要条件。

我的反序列化语法如下所示：

[DataContract(Name="Entity", Namespace="")]
[XmlSerializerFormat]
[KnownType(typeof(SRTSRField))]
[KnownType(typeof(SRTSRValue))]
public class StartRunTestSetResponse
{
    [DataMember(Name="Type"), XmlAttribute("Type")]
    public string type { get; set; }

    [DataMember(Name = "Fields", IsRequired = true), XmlElement("Fields")]
    public List<SRTSRField> fields { get; set; }

    internal StartRunTestSetResponse() { fields = new List<SRTSRField>(); }
}
[DataContract(Name = "Field", Namespace = "")]
[KnownType(typeof(SRTSRValue))]
public class SRTSRField
{
    [DataMember(Name = "Name"), XmlAttribute("Name")]
    public string name {get; set;}

    [DataMember(Name = "Value"), XmlElement("Value")]
    public SRTSRValue value { get; set; }
}
[DataContract(Name = "Value", Namespace = "")]
public class SRTSRValue
{
    [DataMember, XmlText]
    public string value { get; set; }
}

现在，它不起作用。 目前它解析为Fields元素，然后它的任何子级都为null 。

Answer 1

您可以简化模型

public class Entity
{
    [XmlAttribute]
    public string Type { get; set; }
    [XmlArrayItem("Field")]
    public Field[] Fields { get; set; }
}

public class Field
{
    [XmlAttribute]
    public string Name { get; set; }
    public string Value { get; set; }
}

所以反序列化将是

XmlSerializer ser = new XmlSerializer(typeof(Entity));
using (StringReader sr = new StringReader(xmlstring))
{
    var entity = (Entity)ser.Deserialize(sr);

}

Answer 2

[XmlRoot(ElementName="Field")]
public class Field {
    [XmlElement(ElementName="Value")]
    public string Value { get; set; }
    [XmlAttribute(AttributeName="Name")]
    public string Name { get; set; }
}

[XmlRoot(ElementName="Fields")]
public class Fields {
    [XmlElement(ElementName="Field")]
    public List<Field> Field { get; set; }
}

[XmlRoot(ElementName="Entity")]
public class Entity {
    [XmlElement(ElementName="Fields")]
    public Fields Fields { get; set; }
    [XmlAttribute(AttributeName="Type")]
    public string Type { get; set; }
}

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Answer 3

我将使用xml linq创建字典。

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;

namespace ConsoleApplication74
{
    class Program
    {
        const string FILENAME = @"c:\temp\test.xml";
        static void Main(string[] args)
        {
            XDocument doc = XDocument.Load(FILENAME);

            Dictionary<string,int?> dict1 = doc.Descendants("Field")
                .GroupBy(x => (string)x.Attribute("Name"), y => string.IsNullOrEmpty((string)y.Element("Value")) ? null : (int?)y.Element("Value"))
                .ToDictionary(x => x.Key, y => y.FirstOrDefault());


        }

    }



}