如何编程生成一个类型的XML模式？ [英] How do I programmatically generate an xml schema from a type?

问题描述

我想产生一个xs：架构从任何.NET类型编程。我知道我可以使用反射和迭代的公共属性生成它，但有一个内置的方式？

例如：

  [Serializable接口]
公共类Person
{
    [的XmlElement（ISNULLABLE =假）]公共字符串名字{获得;组; }
    [的XmlElement（ISNULLABLE =假）]公共字符串名字{获得;组; }
    [的XmlElement（ISNULLABLE = TRUE）]公共字符串PHONENO {获得;组; }
}
 

所需的输出：

 ＆LT; XS：模式的xmlns：XS =htt​​p://www.w3.org/2001/XMLSchema＆GT;
  ＆LT; XS：元素的名称=人型=人/＆GT;
  ＆LT; XS：复杂类型的名称=人＆GT;
    ＆LT; XS：序列＆GT;
      ＆LT; XS：元素的minOccurs =0的maxOccurs =1的形式=不合格NAME =姓类型=XS：字符串/＆GT;
      ＆LT; XS：元素的minOccurs =0的maxOccurs =1的形式=不合格NAME =姓氏类型=XS：字符串/＆GT;
      ＆LT; XS：元素的minOccurs =0的maxOccurs =1的形式=不合格NAME =PHONENO类型=XS：字符串/＆GT;
    ＆LT; / XS：序列＆GT;
  ＆LT; / XS：复杂类型＆GT;
＆LT; / XS：模式＆GT;
 

解决方案

所以这个作品，我想这是不是因为丑，因为它似乎是：

  VAR SOA preflectionImporter =新的SOA preflectionImporter（）;
VAR xmlTypeMapping = SOA preflectionImporter.ImportTypeMapping（typeof运算（人））;
VAR xmlSchemas =新XmlSchemas（）;
VAR XMLSCHEMA =新的XmlSchema（）;
xmlSchemas.Add（XML模式）;
VAR xmlSchemaExporter =新XmlSchemaExporter（xmlSchemas）;
xmlSchemaExporter.ExportTypeMapping（xmlTypeMapping）;
 

我仍然希望有一个2线的解决方案在那里，好像应该有，感谢小费@dtb

---

修改 只是踢，这里的2行版本（自去precating幽默）

  VAR typeMapping =新的SOA preflectionImporter（）ImportTypeMapping（typeof运算（人））。
新XmlSchemaExporter（新XmlSchemas {新的XmlSchema（）}）ExportTypeMapping（typeMapping）。
 

I'm trying to generate an xs:schema from any .net Type programmatically. I know I could use reflection and generate it by iterating over the public properties, but is there a built in way?

Example:

[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; }
}

Desired Output:

<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>

解决方案

So this works, I guess it wasn't as ugly as it seemed:

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);

I was still hoping there was a 2 line solution out there, it seems like there should be, thanks for the tip @dtb

---

EDIT Just for kicks, here's the 2 line version (self deprecating humor)

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

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