c#序列化xsi:类型和xsd。

I have a schema defined as follows:

我有一个模式定义如下:

<?xml version="1.0" encoding="utf-8"?>
<xs:schema id="Books" xmlns="" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:msdata="urn:schemas-microsoft-com:xml-msdata">
  <xs:element name="Books" msdata:IsDataSet="true" msdata:Locale="en-US">
    <xs:complexType>
      <xs:choice minOccurs="0" maxOccurs="unbounded">
        <xs:element name="Book" type="MyBookType"></xs:element>
      </xs:choice>
    </xs:complexType>
  </xs:element>

  <xs:complexType name="MyBookType">
  ...
  </xs:complexType>

</xs:schema>

Using this schema and xsd.exe, I generate the classes that will be used during serialization. The class generated by the above schema produces the following xml when serialized:

使用这个模式和xsd。exe，我生成将在序列化期间使用的类。上述模式生成的类在序列化时生成以下xml:

<Books>
  <Book>
  ...
  </Book>
</Books>

This xml is used in a SOAP request and the service on the other end expects the following xml:

该xml用于SOAP请求，而另一端的服务期望以下xml:

<Books>
  <Book xsi:type="MyBookType">
  ...
  </Book>
</Books>

How can I edit my schema so that the xsi:type attribute is included in the serialized xml?

如何编辑模式，使xsi:type属性包含在序列化的xml中?

1 个解决方案

#1

Use a derived type, and an XmlInclude attribute. For example:

使用派生类型和XmlInclude属性。例如:

public class Book
{
    public string Title;
    public string Author;
}

public class MyBookType : Book { }

[XmlInclude(typeof(MyBookType))]
[XmlRoot("Books")]
public class Books : List<Book> { }

public void Run()
{
    var b =  new Books();
    b.Add(new MyBookType
        {
            Title = "The Art of War",
            Author = "Sun Tzu"
        });
    b.Add(new MyBookType
        {
            Title = "Great Expectations",
            Author = "Charles Dickens"
        });

    var s = new XmlSerializer(typeof(Books));
    s.Serialize(Console.Out, b);
}

Running this produces this output:

运行此命令将产生以下输出:

<?xml version="1.0" encoding="IBM437"?>
<Books xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
       xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Book xsi:type="MyBookType">
    <Title>The Art of War</Title>
    <Author>Sun Tzu</Author>
  </Book>
  <Book xsi:type="MyBookType">
    <Title>Great Expectations</Title>
    <Author>Charles Dickens</Author>
  </Book>
</Books>

Since you're using a SOAP request, I'm assuming ASMX, which means the serialization happens implicitly. You will need to apply [XmlInclude] to whatever holds the collection of books. This could be a parameter in a webmethod, for example.

因为您正在使用SOAP请求，所以我假设是ASMX，这意味着序列化是隐式的。您将需要将[XmlInclude]应用到包含图书集合的任何地方。例如，这可以是webmethod中的一个参数。

You can automatically generate the appropriate XmlInclude attribute, starting from XSD and WSDL, if you define the types in XSD, with the inheritance relationship I illustrated in C# code.

如果您在XSD中定义了类型，您可以从XSD和WSDL开始自动生成适当的XmlInclude属性，继承关系如c#代码中所示。

In WSDL, the request message might take a Books type, which is a collection of Book. Separately, define a MyBookType which derives from Book but does not extend it.

在WSDL中，请求消息可能采用图书类型，这是图书的集合。另外，定义一个MyBookType，它派生自Book，但不扩展它。

#1