当我的所有.XSD存储为资源时,如何解析.XSD的schemaLocation属性？

Question

我正在开发一个项目,我需要根据嵌套的XSD生成XML文件.例如,ORDER引用了PERSON,PERSON引用了ADDRESS等.

我正在创建一个`XmlReaderSettings'实例来验证XSD,并在生成后验证XML.

我已将XSD作为资源添加到我的程序集中.然后,我XmlSchema为每个资源创建一个实例,从最低到最高,并将其添加到XmlReaderSettings.Schemas集合中.

但是,尝试添加引用另一个架构的架构时失败.我得到一个XmlSchemaException:"对于元素声明,必须存在name或ref属性."

我在下面列出了示例XSD和源代码:

ADDRESS.xsd - 由PERSON.xsd引用

<?xml version="1.0" encoding="utf-8"?>
<xs:schema id="ADDRESS.xsd" attributeFormDefault="unqualified" elementFormDefault="qualified" xmlns:xs="http://www.w3.org/2001/XMLSchema">
  <xs:element name="ADDRESS" >
    <xs:complexType>
      <xs:sequence>
        <xs:element name="ADDRESS1" type="xs:string"/>
        <xs:element name="ADDRESS2" type="xs:string"/>
        <xs:element name="CITY" type="xs:string"/>
        <xs:element name="STATE" type="xs:string"/>
        <xs:element name="ZIP" type="xs:string"/>
      </xs:sequence>
    </xs:complexType>
  </xs:element>
</xs:schema>

PERSON.xsd

<?xml version="1.0" encoding="utf-8"?>
<xs:schema id="PERSON.xsd" attributeFormDefault="unqualified" elementFormDefault="qualified" xmlns:xs="http://www.w3.org/2001/XMLSchema">
  <xs:include schemaLocation="ADDRESS.xsd"/>
  <xs:element name="PERSON" >
    <xs:complexType>
      <xs:sequence>
        <xs:element name="L_NAME" type="xs:string"/>
        <xs:element name="F_NAME" type="xs:string"/>
        <xs:element name="Addresses">
          <xs:complexType>
            <xs:sequence>
              <xs:element ref="ADDRESS" minOccurs="0" maxOccurs="unbounded"/>
            </xs:sequence>
          </xs:complexType>
        </xs:element>
      </xs:sequence>
    </xs:complexType>
  </xs:element>
</xs:schema>

示例代码 - 加载并验证XSD

using System;
using System.IO;
using System.Text;
using System.Xml;
using System.Xml.Schema;
using SchemaTest.Properties;

namespace SchemaTest
{
    class Program
    {
        static void Main(string[] args)
        {
            // create validation settings instance
            var xmlReaderSettings = new XmlReaderSettings
            {
                ValidationType = ValidationType.Schema,
                ValidationFlags = XmlSchemaValidationFlags.ProcessInlineSchema |
                                  XmlSchemaValidationFlags.ProcessSchemaLocation |
                                  XmlSchemaValidationFlags.ReportValidationWarnings
            };

            xmlReaderSettings.ValidationEventHandler += SchemaValidationEventHandler;

            // added XmlResourceResolver, per the accepted answer below
            xMLReaderSettings.Schemas.XmlResolver = new XmlResourceResolver();

            // load schemas into validation settings instance
            LoadSchema(xmlReaderSettings, Resources.PERSON);

            // validate schemas
            xmlReaderSettings.Schemas.Compile();

            Console.WriteLine("Press any key to continue...");
            Console.ReadKey();
        }

        private static void LoadSchema(XmlReaderSettings xmlReaderSettings, string schemaString)
        {
            using (var schemaStream = new MemoryStream(Encoding.Default.GetBytes(schemaString)))
            {
                var schema = XmlSchema.Read(schemaStream, null);

                try
                {
                    xmlReaderSettings.Schemas.Add(schema);
                }
                catch (Exception ex)
                {
                    Console.WriteLine("EXCEPTION: {0}", ex.Message);
                }
            }
        }

        private static void SchemaValidationEventHandler(object sender, ValidationEventArgs e)
        {
            Console.WriteLine("{0}: {1}", e.Severity, e.Message);
        }
    }
}

我不确定这是否与此有关ValidationFlags,但我尝试删除XmlSchemaValidationFlags.ProcessSchemaLocation并仍然收到相同的错误.我也试过传递SchemaValidationEventHandler给XmlSchema.Read方法(而不是示例代码中的null),并且模式似乎被创建好了,但是在将attemtping添加到XmlReaderSettings.Schemas集合时它仍然会引发异常.

编辑 - 2011.11.03

我通过创建我自己的XmlUrlResolver后代来解决这个问题(双关语)XmlResourceResolver,基于XmlResolver下面接受的答案中的类示例.

Answer 1

要消除错误,请在PERSON.xsd文件中将type ="Addresses"更改为name ="Addresses".

最好使用下面的代码来编译模式.重要的是确保建立基础uri(因为您正在从流中读取),使用XmlSchemaSet和自定义解析器(将文件作为嵌入资源读取).

using System;
using System.IO;
using System.Text;
using System.Xml;
using System.Xml.Schema;

namespace ConsoleApplication3
{
    class Program
    {
        class XmlResolver : XmlUrlResolver
        {
            internal const string BaseUri = "schema://";

            public override object GetEntity(Uri absoluteUri, string role, Type ofObjectToReturn)
            {
                if (absoluteUri.Scheme == "schema")
                {
                    switch (absoluteUri.LocalPath)
                    {
                        case "/ADDRESS.xsd":
                            return new MemoryStream(Encoding.UTF8.GetBytes(Resource.ADDRESS));
                        case "/PERSON.xsd":
                            return new MemoryStream(Encoding.UTF8.GetBytes(Resource.PERSON));
                    }
                }
                return base.GetEntity(absoluteUri, role, ofObjectToReturn);
            }
        }

        static void Main(string[] args)
        {
            using (XmlReader reader = XmlReader.Create(new StringReader(Resource.PERSON), new XmlReaderSettings(), XmlResolver.BaseUri))
            {
                XmlSchemaSet xset = RetrieveSchemaSet(reader);
                Console.WriteLine(xset.IsCompiled);
            }
        }

        private static XmlSchemaSet RetrieveSchemaSet(XmlReader reader)
        {
            XmlSchemaSet xset = new XmlSchemaSet() { XmlResolver = new XmlResolver() };
            xset.Add(XmlSchema.Read(reader, null));
            xset.Compile();
            return xset;
        }
    }
}