序列化作为空字符串的 XmlAttribute [关闭]

Question

我有一个代表足球运动员的类​​：

public class PlayerExtended
{
    [XmlAttribute("id")] public string Id { get; set; }
    [XmlAttribute("shortName")] public string ShortName { get; set; }
    [XmlAttribute("firstName")] public string FirstName { get; set; }
    [XmlAttribute("surName")] public string SurName { get; set; }
    [XmlAttribute("shirtNumber")] public string ShirtNumber { get; set; }
    [XmlAttribute("actions")] public string Actions { get; set; }
    [XmlAttribute("substitude")] public string Substitude { get; set; }
    [XmlAttribute("grade")] public string Grade { get; set; }
    [XmlAttribute("iconSmall")] public string IconSmall { get; set; }
    [XmlAttribute("position")] public string Position { get; set; }
    [XmlAttribute("squadPositionID")] public string SquadPositionId { get; set; }
    [XmlAttribute("squadPosition")] public string SquadPosition { get; set; }
    [XmlAttribute("inMinute")] public string InMinute { get; set; }
    [XmlAttribute("outMinute")] public string OutMinute { get; set; }
    [XmlAttribute("captain")] public string Captain { get; set; }
}

在为属性赋值后，其中一个玩家看起来像这样：

属性“Actions”是一个空字符串（NOT NULL）。

如果我序列化它，它看起来像这样：

<player id="51641" shortName="Bürki" firstName="Roman" surName="Bürki" shirtNumber="1" substitude="starter" grade="2,5" iconSmall="xxx.whatever.com" position="11" squadPositionID="1" squadPosition="Torwart"/>

但我希望它看起来像这样：

<player id="51641" shortName="Bürki" firstName="Roman" surName="Bürki" shirtNumber="1" actions="" substitude="starter" grade="2,5" iconSmall="xxx.whatever.com" position="11" squadPositionID="1" squadPosition="Torwart"/>

那么如何序列化一个空字符串的 XmlAttribute 呢？

Answer 1

您是如何生成 XML 的？我似乎无法重现您的问题。

    public class Program
    {
        public static void Main(string[] args)
        {
            using var writer = new StringWriter();
            var serializer = new XmlSerializer(typeof(Player));
            serializer.Serialize(writer, new Player { Name = "", Age = 25 });
            Console.WriteLine(writer);
        }
    }

    public class Player
    {
        [XmlAttribute("name")]
        public string Name { get; set; }


        [XmlAttribute("age")]
        public int Age { get; set; }
    }

上面的代码以您想要的格式生成 name 属性 (name="")。让我知道这个答案对你来说是否足够。

Answer 2

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

// Runtime Target = .NET Core v2.1 or .NET Core v3.1

namespace XmlSerialize
{
    class Program
    {
        static void Main(string[] args)
        {
            var mickey = new Employee { FirstName = "Mickey", LastName = "Mouse" };
            var asterix = new Employee { FirstName = "Asterix", LastName = "" };
            var obelix = new Employee { FirstName = "Obelix", LastName = null };
            var nixnix = new Employee { FirstName = null, LastName = null };

            Console.WriteLine(SerializeXml(mickey) + SerializeXml(asterix) + SerializeXml(obelix) + SerializeXml(nixnix));
        }

        public static string SerializeXml<T>(T instanceToSerialize)
        {
            var serializer = new XmlSerializer(instanceToSerialize.GetType(), string.Empty);
            var result = string.Empty;
            using (var stringWriter = new StringWriter())
            {
                serializer.Serialize(stringWriter, instanceToSerialize);
                result = stringWriter.ToString();
            }
            return result;
        }
    }

    [XmlRoot("Employee")]
    public sealed class Employee
    {
        [XmlAttribute("FirstName")]
        public string FirstName { get; set; }

        [XmlIgnore]
        public string LastName { get; set; }

        [XmlAttribute("LastName")]
        public string SerializableLastName // <------------ Might this help?
        {
            get { return this.LastName ?? string.Empty; }
            set { this.LastName = value; }
        }

        [XmlElement]
        public List<string> Skills { get; set; }
    }
}

输出

<?xml version="1.0" encoding="utf-16"?>
<Employee FirstName="Mickey" LastName="Mouse" />
<Employee FirstName="Asterix" LastName="" />
<Employee FirstName="Obelix" LastName="" />
<Employee LastName="" />

Answer 3

将属性值设置为string.Empty 即可。我正在使用XmlSerializer 将对象转换为 XML。如果我将属性设置为string.Empty，这将导致 XML 中的属性为空。这是一个例子

public class TestClass
{
    [XmlAttribute("test1")]
    public string test1 { get; set; }
    [XmlAttribute("test2")]
    public string test2 { get; set; }
}

var dd = new List<TestClass>();

dd.Add( new TestClass() { test1 = "asdf", test2 = string.Empty }); //will generate empty attribute for test2
dd.Add( new TestClass() { test1 = "asdf" }); //the attribute test2 will be ignored

using (var stringwriter = new System.IO.StringWriter())
{
   var serializer = new XmlSerializer(dd.GetType());

   serializer.Serialize(stringwriter, dd);
   Console.WriteLine( stringwriter.ToString());
}

输出

<?xml version="1.0" encoding="utf-16"?>
<ArrayOfTestClass xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
 xmlns:xsd="http://www.w3.org/2001/XMLSchema">
 <TestClass test1="asdf" test2="" />
 <TestClass test1="asdf" />
</ArrayOfTestClass>