3

最初的问题

我想动态更改类的serializeXmlType 所以我不能使用属性标签来做到这一点)List<IXmlSerializable>IXmlSerializable

我尝试使用XmlAttributeOverrides来做到这一点,但到目前为止没有成功。

这是说明问题的示例代码:
IXmlSerializable 类(来自MSDN):

public class Person : IXmlSerializable
{
    // Private state
    private string personName;


    // Constructors
    public Person(string name)
    {
        personName = name;
    }

    public Person()
    {
        personName = null;
    }


    // Xml Serialization Infrastructure
    public void WriteXml(XmlWriter writer)
    {
        writer.WriteString(personName);
    }

    public void ReadXml(XmlReader reader)
    {
        personName = reader.ReadString();
    }

    public XmlSchema GetSchema()
    {
        return (null);
    }


    // Print
    public override string ToString()
    {
        return (personName);
    }
}

测试类(使用控制台输出):

class Program
{
    static void Main(string[] args)
    {
        List<Person> lPersonList = new List<Person> {
            new Person("First"),
            new Person("Second"),
            new Person("Third")
        };
        XmlAttributeOverrides lOverrides = new XmlAttributeOverrides();
        XmlAttributes lAttributes = new XmlAttributes { XmlType = new XmlTypeAttribute("Employee") };
        lOverrides.Add(typeof(Person), lAttributes);

        XmlSerializer lSerialiser = new XmlSerializer(typeof(List<Person>), lOverrides, null, new XmlRootAttribute("Employees"), null);
        XmlSerializerNamespaces lNamespaces = new XmlSerializerNamespaces();
        lNamespaces.Add("", "");
        lSerialiser.Serialize(Console.Out, lPersonList, lNamespaces);

        System.Console.WriteLine("Enter any key to close.");
        System.Console.ReadKey();
    }
}

这是我想要得到的:

<Employees>
<Employee>First</Employee>
<Employee>Second</Employee>
<Employee>Third</Employee>
</Employees>

但我在运行时收到此错误:

System.InvalidOperationException:只能为 Person 类型指定 XmlRoot 属性。请使用 XmlSchemaProviderAttribute 指定架构类型。

注意 当我的Person 类没有实现IXmlSerializable时,一切正常......

有人可以帮我吗?


选择的解决方案(基于@dbc answer

正如@dbc 指出的那样,使用“代理”类是做我想做的最简单的方法。但正如我所说,我需要动态更改 Person 类型,这意味着我不能使用属性标签。
所以我仍然XmlAttributeOverrides在我的最终设计中使用,这里是:

代理List<Person>(与没有属性标签的@dbc 相同):

public class EmployeesListSurrogate
{
    public List<Person> EmployeeList { get; set; }

    public static implicit operator List<Person>(EmployeesListSurrogate surrogate)
    {
        return surrogate == null ? null : surrogate.EmployeeList;
    }

    public static implicit operator EmployeesListSurrogate(List<Person> employees)
    {
        return new EmployeesListSurrogate { EmployeeList = employees };
    }
}

使用 surrogate 测试类:

class Program
{
    static void Main(string[] args)
    {
        List<Person> lPersonList = new List<Person> {
            new Person("First"),
            new Person("Second"),
            new Person("Third")
        };

        XmlAttributeOverrides lOverrides = new XmlAttributeOverrides();
        XmlAttributes lEmployeesListAttributes = new XmlAttributes { XmlRoot = new XmlRootAttribute("Employees") };
        lOverrides.Add(typeof(EmployeesListSurrogate), lEmployeesListAttributes);
        XmlAttributes lEmployeeAttributes = new XmlAttributes { XmlElements = { new XmlElementAttribute("Employee") } };
        lOverrides.Add(typeof(EmployeesListSurrogate), "EmployeeList", lEmployeeAttributes);

        XmlSerializer lSerializer = new XmlSerializer(typeof(EmployeesListSurrogate), lOverrides);
        XmlSerializerNamespaces lNamespaces = new XmlSerializerNamespaces();
        lNamespaces.Add("", "");
        lSerializer.Serialize(Console.Out, (EmployeesListSurrogate)lPersonList, lNamespaces);
    }
}

最后,我非常感谢@dbc,您的回答非常有帮助且内容丰富,我学到了很多东西。我不能投票给你,但我希望社区会这样做!

4

1 回答 1

3

获取所需 XML 的最简单方法是序列化“代理”类,如下所示:

[XmlRoot("Employees")]
public class EmployeesListSurrogate
{
    [XmlElement("Employee")]
    public List<Person> EmployeeList { get; set; }

    public static implicit operator List<Person>(EmployeesListSurrogate surrogate)
    {
        return surrogate == null ? null : surrogate.EmployeeList;
    }

    public static implicit operator EmployeesListSurrogate(List<Person> employees)
    {
        return new EmployeesListSurrogate { EmployeeList = employees };
    }
}

这完全消除了对XmlAttributeOverrides. 或者您可以使用XmlAttributeOverrideswithXmlAttributes.XmlElementsEmployeeList动态指定 XML 名称。

话虽如此,InvalidOperationException当您尝试应用于[XmlType]也实现的类型时抛出的原因IXmlSerializableXmlSerializer需要通过完全不同的机制返回类型名称,即在属性XmlSchemaProviderAttribute.MethodName中指定的方法。[XmlSchemaProvider]

[XmlSchemaProvider]应用于IXmlSerializable类型时,XmlSerializer将查找该类型的公共静态方法,其名称在属性构造函数中指定并具有以下签名:

    public static XmlQualifiedName GetSchemaMethod(XmlSchemaSet xs)
    {
    }

这种方法的目的有两个:

  1. XmlSchemaSet在序列化类型的实例时,它应该用预期的模式填充。通过测试,我发现它必须填充一些有效的 . 它不能只留空,否则将引发异常。

    (我不知道XmlSerializer在序列化时实际验证架构的程度。通过导出架构信息时也会调用该方法xsd.exe。)

  2. 它应该返回该类型的 XML 类型名称。

    这似乎是 Microsoft 抛出您所看到的异常的原因:因为架构属性提供程序应该返回类型名称,所以XmlType属性会发生冲突。

Person因此,如果我按如下方式修改您的课程:

[XmlSchemaProvider("GetSchemaMethod")]
public class Person : IXmlSerializable
{
    // Private state
    private string personName;

    // Constructors
    public Person(string name)
    {
        personName = name;
    }

    public Person()
    {
        personName = null;
    }

    // This is the method named by the XmlSchemaProviderAttribute applied to the type.
    public static XmlQualifiedName GetSchemaMethod(XmlSchemaSet xs)
    {
        string EmployeeSchema = @"<?xml version=""1.0"" encoding=""utf-16""?>
<xs:schema elementFormDefault=""qualified"" xmlns:xs=""http://www.w3.org/2001/XMLSchema"">
  <xs:import namespace=""http://www.w3.org/2001/XMLSchema"" />
  <xs:element name=""Employee"" nillable=""true"" type=""Employee"" />
  <xs:complexType name=""Employee"" mixed=""true"">
  <xs:sequence>
    <xs:any />
  </xs:sequence>
  </xs:complexType>
</xs:schema>";

        using (var textReader = new StringReader(EmployeeSchema))
        using (var schemaSetReader = System.Xml.XmlReader.Create(textReader))
        {
            xs.Add("", schemaSetReader);
        }
        return new XmlQualifiedName("Employee");
    }

    // Xml Serialization Infrastructure
    public void WriteXml(XmlWriter writer)
    {
        writer.WriteString(personName);
    }

    public void ReadXml(XmlReader reader)
    {
        reader.MoveToContent();
        var isEmpty = reader.IsEmptyElement;
        reader.ReadStartElement();
        if (!isEmpty)
        {
            personName = reader.ReadContentAsString();
            reader.ReadEndElement();
        }
    }

    public XmlSchema GetSchema()
    {
        return (null);
    }

    // Print
    public override string ToString()
    {
        return (personName);
    }
}

并将您的序列化为List<Person>XML,我得到以下结果:

<ArrayOfEmployee xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <Employee>First</Employee>
    <Employee>Second</Employee>
    <Employee>Third</Employee>
</ArrayOfEmployee>

如您所见,Person已成功指定 XML 类型名称。

但是,您希望动态覆盖Personvia的 XML 类型名称,XmlAttributeOverrides而不是将其设置为编译类型。为此,似乎有必要指定一个XmlSchemaProviderAttributeinside XmlAttributes。不幸的是,在里面找不到任何XmlSchemaProvider财产XmlAttributes 微软似乎从未实现过这样的功能。因此,如果您想进一步追求这种设计,您将需要做一些笨拙的事情:GetSchemaMethod()在创建覆盖序列化程序时临时覆盖返回。要记住两件事:

  1. 在引擎盖下,XmlSerializer通过创建动态程序集来工作。如果您构造一个XmlSerializerwith new XmlSerializer(Type)or new XmlSerializer(Type, String),那么 .Net 将缓存该程序集并在第二次构造序列化程序时重用它。

    GetSchemaMethod()因此,在使用其中任何一个构造序列化程序时尝试临时覆盖返回将失败或产生意外结果。

  2. 否则动态程序集不会被缓存,因此您的代码必须手动缓存序列化程序,否则会出现严重的资源泄漏。请参阅使用 StreamReader 和 XmlSerializer 的内存泄漏

    在这些情况下,暂时覆盖GetSchemaMethod()可能工作的返回。

综上所述,以下生成您需要的 XML:

[XmlSchemaProvider("GetSchemaMethod")]
public class Person : IXmlSerializable
{
    // Private state
    private string personName;

    // Constructors
    public Person(string name)
    {
        personName = name;
    }

    public Person()
    {
        personName = null;
    }

    [ThreadStatic]
    static string personXmlTypeName;

    const string defaultXmlTypeName = "Person";

    static string PersonXmlTypeName
    {
        get
        {
            if (personXmlTypeName == null)
                personXmlTypeName = defaultXmlTypeName;
            return personXmlTypeName;
        }
        set
        {
            personXmlTypeName = value;
        }
    }

    public static IDisposable PushXmlTypeName(string xmlTypeName)
    {
        return new PushValue<string>(xmlTypeName, () => PersonXmlTypeName, val => PersonXmlTypeName = val);
    }

    // This is the method named by the XmlSchemaProviderAttribute applied to the type.
    public static XmlQualifiedName GetSchemaMethod(XmlSchemaSet xs)
    {
        string EmployeeSchemaFormat = @"<?xml version=""1.0"" encoding=""utf-16""?>
            <xs:schema elementFormDefault=""qualified"" xmlns:xs=""http://www.w3.org/2001/XMLSchema"">
              <xs:import namespace=""http://www.w3.org/2001/XMLSchema"" />
              <xs:element name=""{0}"" nillable=""true"" type=""{0}"" />
              <xs:complexType name=""{0}"" mixed=""true"">
              <xs:sequence>
                <xs:any />
              </xs:sequence>
              </xs:complexType>
            </xs:schema>";
        var EmployeeSchema = string.Format(EmployeeSchemaFormat, PersonXmlTypeName);

        using (var textReader = new StringReader(EmployeeSchema))
        using (var schemaSetReader = System.Xml.XmlReader.Create(textReader))
        {
            xs.Add("", schemaSetReader);
        }
        return new XmlQualifiedName(PersonXmlTypeName);
    }

    // Xml Serialization Infrastructure
    public void WriteXml(XmlWriter writer)
    {
        writer.WriteString(personName);
    }

    public void ReadXml(XmlReader reader)
    {
        reader.MoveToContent();
        var isEmpty = reader.IsEmptyElement;
        reader.ReadStartElement();
        if (!isEmpty)
        {
            personName = reader.ReadContentAsString();
            reader.ReadEndElement();
        }
    }

    public XmlSchema GetSchema()
    {
        return (null);
    }

    // Print
    public override string ToString()
    {
        return (personName);
    }
}

public struct PushValue<T> : IDisposable
{
    Action<T> setValue;
    T oldValue;

    public PushValue(T value, Func<T> getValue, Action<T> setValue)
    {
        if (getValue == null || setValue == null)
            throw new ArgumentNullException();
        this.setValue = setValue;
        this.oldValue = getValue();
        setValue(value);
    }

    #region IDisposable Members

    // By using a disposable struct we avoid the overhead of allocating and freeing an instance of a finalizable class.
    public void Dispose()
    {
        if (setValue != null)
            setValue(oldValue);
    }

    #endregion
}

public static class PersonEmployeeListSerializerFactory
{
    static Dictionary<Tuple<string, string>, XmlSerializer> serializers;
    static object padlock = new object();

    static PersonEmployeeListSerializerFactory()
    {
        serializers = new Dictionary<Tuple<string, string>, XmlSerializer>();
    }

    public static XmlSerializer GetSerializer(string rootName, string personName)
    {
        lock (padlock)
        {
            XmlSerializer serializer;
            var key = Tuple.Create(rootName, personName);
            if (!serializers.TryGetValue(key, out serializer))
            {
                using (Person.PushXmlTypeName(personName))
                {
                    var lOverrides = new XmlAttributeOverrides();
                    //var lAttributes = new XmlAttributes();
                    //lOverrides.Add(typeof(Person), lAttributes);

                    serializers[key] = serializer = new XmlSerializer(typeof(List<Person>), lOverrides, new Type[0], new XmlRootAttribute(rootName), null);
                }
            }
            return serializer;
        }
    }
}

然后做

var lSerialiser = PersonEmployeeListSerializerFactory.GetSerializer("Employees", "Employee");

var lNamespaces = new XmlSerializerNamespaces();
lNamespaces.Add("", "");

var sb = new StringBuilder();
using (var writer = new StringWriter(sb))
    lSerialiser.Serialize(writer, lPersonList, lNamespaces);

Console.WriteLine(sb);

但是正如您所看到的,这比使用最初显示的代理要复杂得多。

显示两个选项的示例小提琴。

于 2016-07-02T09:02:28.493 回答