[英]Serialize array of derived type property of derived type
我有一个简单的.Net framework C#控制台应用程序,该应用程序序列化了一个派生类型的类,其中一个属性也是一个派生类型。
派生类的名称与基类相同,但在不同的名称空间中以防止它们冲突。 似乎XmlSerializer
使用的反射不能很好地解决这个问题。 也许有一些方法可以解决这些属性,但我仍然可以使用漂亮的名称(因为使用时它将是DLL接口)来结束基类,而XML也使用漂亮的名称(因为它将是可人工编辑的)。不需要派生类的漂亮名称(尽管这是一个奖励)。
XML希望如下所示:
<?xml version="1.0" encoding="utf-8"?>
<Person xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Details>
<Detail>
<Description>bald</Description>
</Detail>
<Detail>
<Description>red tie</Description>
</Detail>
</Details>
</Person>
但是我可以毫无例外地得到的最接近的是<Detail>
元素所在的位置
<Detail xsi:type="DerivedDetail"> ... </Detail>
必须添加此xs:type
属性对于人类可编辑的XML并不是最好的选择。
这是通过以下C#代码实现的。 如果我删除了标记的XmlType
属性,则该元素应在没有xsi:type
属性的情况下进行序列化,但会出现异常:
InvalidOperationException:类型'Test.Detail'和'Test.Xml.Detail'都使用来自名称空间''的XML类型名称'Detail'。 使用XML属性为类型指定唯一的XML名称和/或名称空间。
我尝试将派生的Xml.Detail
类标记为匿名XML类型,但是异常显示为:
InvalidOperationException:无法包含匿名类型'Test.Xml.Detail'。
我读过许多类似的问题,但还没有遇到任何可以解决这个问题的东西。
在下面的代码中, Person
是一个抽象类,它具有一个属性,该属性是一个抽象类型Detail
的数组。 这些类型分别由Xml.Person
和Xml.Detail
派生。 该程序创建一个测试Xml.Person
对象并尝试对其进行序列化:
using System;
using System.Text;
using System.IO;
using System.Xml;
using System.Xml.Serialization;
namespace Test
{
class Program
{
static void Main(string[] args)
{
// Create test details array
var TestDetails = new Xml.Detail[]
{
new Xml.Detail
{
Description = "bald"
},
new Xml.Detail
{
Description = "red tie"
}
};
// create test person object that holds details array
var TestBar = new Xml.Person()
{
Details = TestDetails
};
// serialize the person object
var s = new Xml.Serializer();
var TestOutput = s.Serialize(TestBar);
Console.WriteLine(TestOutput);
}
}
// base classes
public abstract class Person
{
public abstract Detail[] Details { get; set; }
}
public abstract class Detail
{
public abstract string Description { get; set; }
}
namespace Xml
{
// derived classes
[Serializable]
[XmlType(AnonymousType = true)]
[XmlRoot(IsNullable = false)]
public class Person : Test.Person
{
[XmlArrayItem("Detail", typeof(Detail))]
[XmlArray(IsNullable = false)]
public override Test.Detail[] Details { get; set; }
}
// This attribute makes serialization work but also adds the xsi:type attribute
[XmlType("DerivedDetail")]
[Serializable]
public class Detail : Test.Detail
{
public override string Description { get; set; }
}
// class that does serializing work
public class Serializer
{
private static XmlSerializer PersonSerializer =
new XmlSerializer(typeof(Person), new Type[] { typeof(Detail) });
public string Serialize(Test.Person person)
{
string Output = null;
var Stream = new MemoryStream();
var Encoding = new UTF8Encoding(false, true);
using (var Writer = new XmlTextWriter(Stream, Encoding))
{
Writer.Formatting = Formatting.Indented;
PersonSerializer.Serialize(Writer, person);
Output = Encoding.GetString(Stream.ToArray());
}
Stream.Dispose();
return Output;
}
}
}
}
不知道没有成员字段时为什么要使用基类而不是接口。 无论如何,我都假定您希望Xml.Person
是抽象Person
或从抽象Person
派生的任何类的具体实例,而不用XML属性装饰抽象Person
。 我通过强制抽象Person
在序列化之前成为Xml.Person
的具体实例来实现这一点。 请用Test
替换XmlSerializationProject
。
using System;
using System.IO;
using System.Text;
using System.Xml;
using System.Xml.Serialization;
namespace XmlSerializationProject
{
class Program
{
static void Main(string[] args)
{
// Create test details array
var TestDetails = new Xml.Detail[]
{
new Xml.Detail
{
Description = "bald"
},
new Xml.Detail
{
Description = "red tie"
}
};
// create test person object that holds details array
var TestBar = new Xml.Person()
{
Details = TestDetails
};
// serialize the person object
var s = new Xml.Serializer();
var TestOutput = s.Serialize(TestBar);
Console.WriteLine(TestOutput);
Console.ReadKey();
}
}
// base classes
public abstract class Person
{
public abstract Detail[] Details { get; set; }
}
public abstract class Detail
{
public abstract string Description { get; set; }
}
namespace Xml
{
[Serializable]
[XmlType(AnonymousType = true)]
[XmlRoot(IsNullable = false)]
public class Person : XmlSerializationProject.Person
{
public Person()
{ }
public Person(XmlSerializationProject.Person person)
{
// Deep copy
if (person.Details == null) return;
this.Details = new Detail[person.Details.Length];
for (int i = 0; i < person.Details.Length; i++)
{
this.Details[i] = new Detail { Description = person.Details[i].Description };
}
}
[XmlArray(ElementName = "Details")]
[XmlArrayItem("Detail", typeof(Detail))]
[XmlArrayItem("ODetail", typeof(XmlSerializationProject.Detail))]
public override XmlSerializationProject.Detail[] Details
{
get;
set;
}
}
[Serializable]
public class Detail : XmlSerializationProject.Detail
{
public override string Description { get; set; }
}
// class that does serializing work
public class Serializer
{
private static readonly XmlSerializer PersonSerializer;
private static Serializer()
{
var xmlAttributeOverrides = new XmlAttributeOverrides();
// Change original "Detail" class's element name to "AbstractDetail"
var xmlAttributesOriginalDetail = new XmlAttributes();
xmlAttributesOriginalDetail.XmlType = new XmlTypeAttribute() { TypeName = "AbstractDetail" };
xmlAttributeOverrides.Add(typeof(XmlSerializationProject.Detail), xmlAttributesOriginalDetail);
// Ignore Person.Details array
var xmlAttributesOriginalDetailsArray = new XmlAttributes();
xmlAttributesOriginalDetailsArray.XmlIgnore = true;
xmlAttributeOverrides.Add(typeof(XmlSerializationProject.Person), "Details", xmlAttributesOriginalDetailsArray);
PersonSerializer = new XmlSerializer(
typeof(Person), xmlAttributeOverrides, new Type[] { typeof(Detail) }, new XmlRootAttribute(), "default");
}
public string Serialize(XmlSerializationProject.Person person)
{
return Serialize(new Person(person));
}
public string Serialize(Person person)
{
string Output = null;
var Stream = new MemoryStream();
var Encoding = new UTF8Encoding(false, true);
using (var Writer = new XmlTextWriter(Stream, Encoding))
{
Writer.Formatting = Formatting.Indented;
PersonSerializer.Serialize(Writer, person);
Output = Encoding.GetString(Stream.ToArray());
}
Stream.Dispose();
return Output;
}
}
}
}
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