将 XML 反序列化为在 C# 中具有对象列表的对象

Question

我有以下 xml，我想将其反序列化为一个对象。

<result>
    <reporttype>2</reporttype>
    <items>
        <item>
            <sku>0B0005</sku>
            <style>0B0005.DAK.GREY</style>
            <reason>Barcode cannot be moved to different SKUs</reason>
        </item>
        <item>
            <sku>0B0006</sku>
            <style>0B0006.DAK.GREY</style>
            <reason>Barcode cannot be moved to different SKUs</reason>
        </item>
    </items>
</result>

但是下面的代码没有填充项目列表，谁能指出我这里做错了什么

string inputString = @"<result><reporttype>2</reporttype><items><item><sku>0B0005</sku><style>0B0005.DAK.GREY</style><reason>Barcode cannot be moved to different SKUs</reason></item><item><sku>0B0005</sku><style>0B0005.DAK.GREY</style><reason>Barcode cannot be moved to different SKUs</reason></item></items></result>";

XmlDocument doc = new XmlDocument();
doc.LoadXml(inputString);

XmlSerializer serializer = new XmlSerializer(typeof(Result));
StringReader rdr = new StringReader(doc.InnerXml);
Result resultingMessage = (Result)serializer.Deserialize(rdr);

public enum ReportType {
    [XmlEnum("0")]
    InternalErrorReport,
    [XmlEnum("1")]
    ErrorReport,
    [XmlEnum("2")]
    InternalSuccessReport
}

[XmlRoot(ElementName = "result")]
public class Result {
    [XmlElement(ElementName = "reporttype")]
    public ReportType reportType { get; set; }
    [XmlElement(ElementName = "items")]
    public List<Item> items = new List<Item>();

    public string error { get; set; }

    public class Item {
        [XmlElement(ElementName = "sku")]
        string sku { get; set; }
        [XmlElement(ElementName = "style")]
        string style { get; set; }
        [XmlElement(ElementName = "reason")]
        string reason { get; set; }
    }
}

或者有更好的方法吗？

Answer 1

items 应该是 Xmlroot 元素并且它包含 XmlElement 项目，你必须告诉它什么时候反序列化到对象。试试这个：

public class Item
{
    [XmlElement(ElementName = "sku")]
    public string Sku { get; set; }
    [XmlElement(ElementName = "style")]
    public string Style { get; set; }
    [XmlElement(ElementName = "reason")]
    public string Reason { get; set; }
}

[XmlRoot(ElementName = "items")]
public class Items
{
    [XmlElement(ElementName = "item")]
    public List<Item> Item { get; set; }
}

[XmlRoot(ElementName = "result")]
public class Result
{
    [XmlElement(ElementName = "reporttype")]
    public string Reporttype { get; set; }
    [XmlElement(ElementName = "items")]
    public Items Items { get; set; }
}

Answer 2

您可以为 items 添加两个属性属性 - 以满足序列化

[XmlRoot(ElementName = "result")]
public class Result 
{
    [XmlArray("items")]
    [XmlArrayItem("item")]
    public List<Item> items = new List<Item>();
}

或者只为项目class设置类型属性(XmlType)。
那么只使用 XmlArray 属性就足够了 Result.items 属性。或者根本不使用任何属性，因为属性的名称与 xml.

中的元素名称匹配

[XmlType("item")]
public class Item 
{
    [XmlElement(ElementName = "sku")]
    public string sku { get; set; }
    [XmlElement(ElementName = "style")]
    public string style { get; set; }
    [XmlElement(ElementName = "reason")]
    public string reason { get; set; }
}

当然还有制作属性public

Answer 3

变量需要 public.

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

namespace ConsoleApplication1
{
    class Program
    {
        const string FILENAME = @"c:\temp\test.xml";
        static void Main(string[] args)
        {
            string xml = File.ReadAllText(FILENAME);

            XmlSerializer serializer = new XmlSerializer(typeof(Result));
            StringReader rdr = new StringReader(xml);
            Result resultingMessage = (Result)serializer.Deserialize(rdr);


        }
    }
    public enum ReportType
    {
        [XmlEnum("0")]
        InternalErrorReport,
        [XmlEnum("1")]
        ErrorReport,
        [XmlEnum("2")]
        InternalSuccessReport
    }

    [XmlRoot(ElementName = "result")]
    public class Result
    {
        [XmlElement(ElementName = "reporttype")]
        public ReportType reportType { get; set; }
        public Items items { get; set; }
        public string error { get; set; }


    }
    [XmlRoot("items")]
    public class Items
    {
        [XmlElement(ElementName = "item")]
        public List<Item> items = new List<Item>();
    }
    [XmlRoot("item")]
    public class Item
    {
        [XmlElement(ElementName = "sku")]
        public string sku { get; set; }
        [XmlElement(ElementName = "style")]
        public string style { get; set; }
        [XmlElement(ElementName = "reason")]
        public string reason { get; set; }
    }
}

Answer 4

如此处所述，您需要将列表标记为 XmlArray，同时指定 XmlArrayItem：Deserializing nested lists with XmlSerializer

所以代码变成：

        string inputString = @"<result><error>error test</error><reporttype>2</reporttype><items><item><sku>0B0005</sku><style>0B0005.DAK.GREY</style><reason>Barcode cannot be moved to different SKUs</reason></item><item><sku>0B0005</sku><style>0B0005.DAK.GREY</style><reason>Barcode cannot be moved to different SKUs</reason></item></items></result>";

        XmlDocument doc = new XmlDocument();
        doc.LoadXml(inputString);

        XmlSerializer serializer = new XmlSerializer(typeof(Result));
        object resultingMessage = null;
        using (StringReader rdr = new StringReader(doc.InnerXml)) {
            resultingMessage = (Result)serializer.Deserialize(rdr);
        }

和类:

public enum ReportType {
    [XmlEnum("0")]
    InternalErrorReport,
    [XmlEnum("1")]
    ErrorReport,
    [XmlEnum("2")]
    InternalSuccessReport
}

[XmlRoot(ElementName = "result")]
public class Result {
    [XmlElement(ElementName = "reporttype")]
    public ReportType reporttype { get; set; }
    [XmlArray("items")]
    [XmlArrayItem("item")]
    public List<Item> items { get; set; }
    [XmlElement(ElementName = "error")]
    public string error { get; set; }

    [XmlRoot(ElementName = "items\item")]
    public class Item {
        [XmlElement(ElementName = "sku")]
        public string sku { get; set; }
        [XmlElement(ElementName = "style")]
        public string style { get; set; }
        [XmlElement(ElementName = "reason")]
        public string reason { get; set; }
    }

}

另请注意，我将字符串 reader 封装在一个 using 块中，以便在读取结束后处理该对象。

将 XML 反序列化为在 C# 中具有对象列表的对象

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