如何从类创建XSD架构？

您可以使用

XSD.exe

public sealed class Configuration{ public string Name { get; set; } public Levels Level { get; set; } public ConfigurationSpec Spec { get; set; }} public abstract class ConfigurationSpec { } public class ConfigurationSpec1    {   }public class ConfigurationSpec2 {   }

结果是

<?xml version="1.0" encoding="utf-8"?><xs:schema elementFormDefault="qualified" xmlns:xs="http://www.w3.org/2001/XMLSchema">  <xs:element name="Levels" type="Levels" />  <xs:simpleType name="Levels">    <xs:restriction base="xs:string">      <xs:enumeration value="Easy" />      <xs:enumeration value="Medium" />      <xs:enumeration value="Hard" />    </xs:restriction>  </xs:simpleType>  <xs:element name="Configuration" nillable="true" type="Configuration" />  <xs:complexType name="Configuration">    <xs:sequence>      <xs:element minOccurs="0" maxOccurs="1" name="Name" type="xs:string" />      <xs:element minOccurs="1" maxOccurs="1" name="Level" type="Levels" />      <xs:element minOccurs="0" maxOccurs="1" name="Spec" type="ConfigurationSpec" />    </xs:sequence>  </xs:complexType>  <xs:complexType name="ConfigurationSpec" abstract="true" />  <xs:element name="ConfigurationSpec" nillable="true" type="ConfigurationSpec" />  <xs:element name="ConfigurationSpec1" nillable="true" type="ConfigurationSpec1" />  <xs:complexType name="ConfigurationSpec1" />  <xs:element name="ConfigurationSpec2" nillable="true" type="ConfigurationSpec2" />  <xs:complexType name="ConfigurationSpec2" /></xs:schema>

您要做的就是编译程序集，并

XSD.exe

XSD.exe /?

例：

XSD.exe C:DevProject1BinDebuglibrary.dll