如何以編程方式從類型生成xml模式？

Question

我試圖以編程方式從任何.net類型生成xs：schema。我知道我可以使用反射並通過遍歷公共屬性來生成它，但是是否有內置方式？

實施例：

[Serializable] 
public class Person 
{ 
    [XmlElement(IsNullable = false)] public string FirstName { get; set; } 
    [XmlElement(IsNullable = false)] public string LastName { get; set; } 
    [XmlElement(IsNullable = true)] public string PhoneNo { get; set; } 
} 

所需的輸出：

<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema"> 
    <xs:element name="Person" type="Person" /> 
    <xs:complexType name="Person"> 
    <xs:sequence> 
     <xs:element minOccurs="0" maxOccurs="1" form="unqualified" name="FirstName" type="xs:string" /> 
     <xs:element minOccurs="0" maxOccurs="1" form="unqualified" name="LastName" type="xs:string" /> 
     <xs:element minOccurs="0" maxOccurs="1" form="unqualified" name="PhoneNo" type="xs:string" /> 
    </xs:sequence> 
    </xs:complexType> 
</xs:schema> 

Answer 1

所以這個作品，我想這是不醜，因爲它似乎是：

var soapReflectionImporter = new SoapReflectionImporter(); 
var xmlTypeMapping = soapReflectionImporter.ImportTypeMapping(typeof(Person)); 
var xmlSchemas = new XmlSchemas(); 
var xmlSchema = new XmlSchema(); 
xmlSchemas.Add(xmlSchema); 
var xmlSchemaExporter = new XmlSchemaExporter(xmlSchemas); 
xmlSchemaExporter.ExportTypeMapping(xmlTypeMapping); 

我仍然希望有一個2線的解決方案在那裏，好像有是應該的，感謝小費@dtb

---

編輯 只是爲了KIC KS，這裏的2行版本（自我貶低的幽默）

var typeMapping = new SoapReflectionImporter().ImportTypeMapping(typeof(Person)); 
new XmlSchemaExporter(new XmlSchemas { new XmlSchema() }).ExportTypeMapping(typeMapping); 

Answer 2

可以編程方式調用xsd.exe：

1. 添加XSD.EXE作爲組件參考。
2. using XsdTool;
3. Xsd.Main(new[] { "myassembly.dll", "/type:MyNamespace.MyClass" });

---

您還可以使用反射來看看XsdTool.Xsd.ExportSchemas方法。它使用公共的XmlReflectionImporter,XmlSchemas,XmlSchemaXmlSchemaExporter和XmlTypeMapping類從.NET類型創建一個模式。

本質上它做到這一點：

var importer = new XmlReflectionImporter(); 
var schemas = new XmlSchemas(); 
var exporter = new XmlSchemaExporter(schemas); 

var xmlTypeMapping = importer.ImportTypeMapping(typeof(Person)); 
exporter.ExportTypeMapping(xmlTypeMapping); 

schemas.Compile(..., false); 

for (var i = 0; i < schemas.Count; i++) 
{ 
    var schema = schemas[i]; 
    schema.Write(...); 
}     ↑ 

您應該能夠通過傳遞合適的作家到XmlSchema.Write方法來定製輸出。

Answer 3

XML模式定義工具從XDR，XML和XSD文件或運行時程序集中的類生成XML模式或公共語言運行時類。

http://msdn.microsoft.com/en-us/library/x6c1kb0s(VS.71).aspx

Answer 4

我相信這是你要找的內容：從上面Writing your own XSD.exe

借款代碼：

using System; 
using System.IO; 
using System.Collections.Generic; 
using System.Reflection; 
using System.Text; 
using System.Xml; 
using System.Xml.Serialization; 
using System.Xml.Schema; 
using System.CodeDom; 
using System.CodeDom.Compiler; 

using Microsoft.CSharp; 

using NUnit.Framework; 

namespace XmlSchemaImporterTest 
{ 
    [TestFixture] 
    public class XsdToClassTests 
    { 
     // Test for XmlSchemaImporter 
     [Test] 
     public void XsdToClassTest() 
     { 
      // identify the path to the xsd 
      string xsdFileName = "Account.xsd"; 
      string path = Path.GetDirectoryName(Assembly.GetExecutingAssembly().Location); 
      string xsdPath = Path.Combine(path, xsdFileName); 

      // load the xsd 
      XmlSchema xsd; 
      using(FileStream stream = new FileStream(xsdPath, FileMode.Open, FileAccess.Read)) 
      { 
       xsd = XmlSchema.Read(stream, null); 
      } 
      Console.WriteLine("xsd.IsCompiled {0}", xsd.IsCompiled); 

      XmlSchemas xsds = new XmlSchemas(); 
      xsds.Add(xsd); 
      xsds.Compile(null, true); 
      XmlSchemaImporter schemaImporter = new XmlSchemaImporter(xsds); 

      // create the codedom 
      CodeNamespace codeNamespace = new CodeNamespace("Generated"); 
      XmlCodeExporter codeExporter = new XmlCodeExporter(codeNamespace); 

      List maps = new List(); 
      foreach(XmlSchemaType schemaType in xsd.SchemaTypes.Values) 
      { 
       maps.Add(schemaImporter.ImportSchemaType(schemaType.QualifiedName)); 
      } 
      foreach(XmlSchemaElement schemaElement in xsd.Elements.Values) 
      { 
       maps.Add(schemaImporter.ImportTypeMapping(schemaElement.QualifiedName)); 
      } 
      foreach(XmlTypeMapping map in maps) 
      { 
       codeExporter.ExportTypeMapping(map); 
      } 

      RemoveAttributes(codeNamespace); 

      // Check for invalid characters in identifiers 
      CodeGenerator.ValidateIdentifiers(codeNamespace); 

      // output the C# code 
      CSharpCodeProvider codeProvider = new CSharpCodeProvider(); 

      using(StringWriter writer = new StringWriter()) 
      { 
       codeProvider.GenerateCodeFromNamespace(codeNamespace, writer, new CodeGeneratorOptions()); 
       Console.WriteLine(writer.GetStringBuilder().ToString()); 
      } 

      Console.ReadLine(); 
     } 

     // Remove all the attributes from each type in the CodeNamespace, except 
     // System.Xml.Serialization.XmlTypeAttribute 
     private void RemoveAttributes(CodeNamespace codeNamespace) 
     { 
      foreach(CodeTypeDeclaration codeType in codeNamespace.Types) 
      { 
       CodeAttributeDeclaration xmlTypeAttribute = null; 
       foreach(CodeAttributeDeclaration codeAttribute in codeType.CustomAttributes) 
       { 
        Console.WriteLine(codeAttribute.Name); 
        if(codeAttribute.Name == "System.Xml.Serialization.XmlTypeAttribute") 
        { 
         xmlTypeAttribute = codeAttribute; 
        } 
       } 
       codeType.CustomAttributes.Clear(); 
       if(xmlTypeAttribute != null) 
       { 
        codeType.CustomAttributes.Add(xmlTypeAttribute); 
       } 
      } 
     } 
    } 
} 

Answer 5

我找到了accepted answer根據我的一些屬性生成了不正確的模式。例如它忽視了標有[XmlEnum(Name="Foo")]

我相信這是正確的方式（您使用XmlSerializer給出）枚舉值自定義名稱，是相當簡單太： http://blogs.msdn.com/b/youssefm/archive/2010/05/13/using-xml-schema-import-and-export-for-xmlserializer.aspx