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如何强制使用xsi:type属性?

xml-serialization .net

Tim Erickson · 技术社区 · 16 年前

如何强制.NET的XmlSerializer添加 xsi:type=“食品类” 到类型的成员/节点 FooClass ?

该场景是当前发布的应用程序,在v.1中:

fooclass继承foobseclass
FoopropertyA在FoobaseClass上
FoopropertyB在Fooclass上
foobseclass装饰有[xmlclude(typeof(fooclass))]
BazClass具有foobaseClass类型的成员foo
所有baz.foo集都指向一个fooclass实例
baz.foo的所有用法都需要foopropertyb(因此fooclass实例与foobaseclass)

目标:完全移除foobseclass,将foobseclass的成员推到fooclass中, 同时保持向后序列化兼容性

问题:然后我丢失了baz.foo序列化上的xsi:type=“fooclass”属性。

换句话说,xmlserializer.serialize输出

public class BazClass
{
    public BazClass()
    {
        Foo = new FooClass { A = 5, B = "Hello" };
    }
    public FooClass Foo { get; set; }
}

public class FooClass
{
    public int FooPropertyA { get; set; }
    public string FooPropertyB { get; set; }
}

需要

<Baz>
    <Foo xsi:type="FooClass">
        <FooPropertyA>Hello</FooPropertyA>
        <FooPropertyB>5</FooPropertyB>
    </Foo>
</Baz>

移除foobasclass很容易,但是xmlserializer不再将xsi:type=“fooclass”放在baz/foo和so v.1 xmlserializer上。反序列化将实例化foobasclass实例,而不是设置fooropertyB,并将其分配给父baz实例的foo属性。因此,任何检查baz.foo是fooclass还是直接强制转换的代码都会失败。

xsi:type属性被自动放入v.1代码中,该代码是

public class BazClass
{
    public BazClass()
    {
        Foo = new FooClass { A = 5, B = "Hello" };
    }
    public FooBaseClass Foo { get; set; }
}

public class FooClass : FooBaseClass
{
    public string FooPropertyB { get; set; }
}

[XmlInclude(typeof(FooClass))]    
public class FooBaseClass
{
    public int FooPropertyA { get; set; }
}

我认为简短的答案是,至少在没有实现I(XML)可序列化或编写自定义序列化代码的情况下是不行的。不过,我愿意接受好的建议。同时我已经实现了 ~~解决办法~~ 下面的黑,我希望有更优雅的东西,或者至少可以让我以某种方式完全删除foobseclass。

BazClass
{
    [XmlElement("Foo")]
    public FooBaseClass XmlFoo { get { return Foo; } set { Foo = (StartPicture)value; } }

    [XmlIgnore]
    public FooClass Foo { get; set; }
}    

FooClass : FooBaseClass
{
    public int FooPropertyB { get; set; }
    public string FooPropertyA { get; set; }
}

[XmlInclude("FooClass")]
FooBaseClass
{
}

3 回复 | 直到 12 年前

Pavel Minaev 16 年前

XmlSerializer 有时可能非常愚蠢和直截了当,在这种情况下,这对您有利。只需手动放置:

public class FooClass
{
    public int FooPropertyA { get; set; }
    public string FooPropertyB { get; set; }

    [XmlAttribute("type", Namespace="http://www.w3.org/2001/XMLSchema-instance")]
    public string XsiType
    {
        get { return "Foo"; }
        set { }
    }
}

John Saunders 16 年前

我在制作以下内容时没有遇到任何问题:

<?xml version="1.0" encoding="utf-8"?>
<BazClass xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
   xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <Foo xsi:type="FooClass">
        <FooPropertyA>Hello</FooPropertyA>
        <FooPropertyB>5</FooPropertyB>
    </Foo>
</BazClass>

从

[XmlInclude(typeof(FooClass))]
//[XmlType(TypeName = "FooBase", Namespace = "urn:namespace", IncludeInSchema = true)]
public class FooBaseClass
{
    public string FooPropertyA { get; set; }
}

//[XmlType(TypeName = "Foo", Namespace = "urn:namespace", IncludeInSchema = true)]
public class FooClass : FooBaseClass
{
    public int FooPropertyB { get; set; } 
}

public class BazClass
{
    public FooBaseClass Foo { get; set; }
}

(注意 XmlType 属性被注释掉。我想看看如果指定了名称空间会发生什么)

请显示您使用的代码及其生成的XML。

Tony Wall 12 年前

要支持其他命名空间中类型的继承,您需要使用类似于pavel minaev建议的解决方案,但使用xmlQualifiedName类型化属性而不是字符串,例如。

using System.Xml;
using System.Xml.Schema;
using System.Xml.Serialization;

namespace Test
{
    /// <summary>
    /// Base class which is XML serializable and extensible.
    /// </summary>
    [XmlRoot(XmlRootName, Namespace = XmlRootNamespace)]
    public abstract class BaseClassInOtherNamespace
    {
        /// <summary>
        /// Name of the XML element 
        /// </summary>
        public const string XmlRootName = "Base";

        /// <summary>
        /// XML namespace in which this type is defined.
        /// </summary>
        public const string XmlRootNamespace = "urn:base";

        /// <summary>
        /// Creates an instance which serializes as the correct inherited XML type.
        /// </summary>
        protected BaseClassInOtherNamespace(XmlQualifiedName xsiType)
        {
            XsiType = xsiType;
        }

        /// <summary>
        /// XML type for serialization.
        /// </summary>
        [XmlAttribute("type", Namespace = XmlSchema.InstanceNamespace)]
        public XmlQualifiedName XsiType { get; set; }

        /// <summary>
        /// Some base property.
        /// </summary>
        public int BaseProperty { get; set; }
    }

    /// <summary>
    /// Inheriting class extending the base class, created in a different XML namespace.
    /// </summary>
    [XmlRoot(XmlRootName, Namespace = XmlRootNamespace)]
    [XmlType(XmlTypeName, Namespace = XmlTypeNamespace)]
    public class InheritingClass : BaseClassInOtherNamespace
    {
        /// <summary>
        /// Name of the XML element 
        /// </summary>
        public const string XmlTypeName = "Inheriting";

        /// <summary>
        /// XML namespace in which this type is defined.
        /// </summary>
        public const string XmlTypeNamespace = "urn:other";

        /// <summary>
        /// Creates an instance.
        /// </summary>
        public InheritingClass() : base(new XmlQualifiedName(XmlTypeName, XmlTypeNamespace))
        {
        }

        /// <summary>
        /// Some new property in a different (inheriting) namespace.
        /// </summary>
        public int InheritingProperty { get; set; }
    }
}

将正确序列化(和反序列化)为:

<Base xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:q1="urn:other" xsi:type="q1:Inheriting" xmlns="urn:base">
  <BaseProperty>0</BaseProperty>
  <q1:InheritingProperty>0</q1:InheritingProperty>
</Base>

这满足了真正可扩展的多态XML类型的要求,即基类可以在任何地方使用,以后的附加组件可以在.NET和XSD验证中正确地分配、序列化和反序列化。

更进一步,您还可以使用xmlnamespacedeclarationsattribute添加一个xmlserializernamespaces属性来指定首选前缀,并删除任何不需要的命名空间,例如xmlns:xsd(我们只使用xmlns:xsi),甚至是用于非继承的xml序列化类的xsi命名空间。