根据多个架构定义验证 XML 文件

Xerces 中的模式内容是 （a） 非常非常迂腐的，并且 （b） 当它不喜欢它找到的东西时，会给出完全无用的错误消息。这是一个令人沮丧的组合。

python 中的模式内容可能更加宽容，并且让模式中的小错误没有报告。

现在，如果如您所说，c.xsd 包含 b.xsd，而 b.xsd 包含 a.xsd，则无需将所有三个都加载到架构工厂中。这不仅没有必要，而且可能会混淆 Xerces 并导致错误，所以这可能是您的问题。只需将 c.xsd 传递给工厂，让它解析 b.xsd 和 a.xsd 本身，它应该相对于 c.xsd 执行此操作。

因此，以防万一其他人在这里遇到同样的问题，我需要从单元测试中加载父架构（和隐式子架构） - 作为资源 - 以验证 XML 字符串。我使用 Xerces XMLSchemFactory 和 Java 6 验证器来做到这一点。

为了通过包含正确加载子架构，我必须编写一个自定义资源解析器。代码可以在这里找到：

https://code.google.com/p/xmlsanity/source/browse/src/com/arc90/xmlsanity/validation/ResourceResolver.java

若要使用解析程序，请在架构工厂中指定它：

xmlSchemaFactory.setResourceResolver(new ResourceResolver());

它将使用它通过类路径解析您的资源（在我的情况下来自 src/main/resources）。欢迎对此发表任何意见...

http://www.kdgregory.com/index.php?page=xml.parsing“单个文档的多个架构”部分'

我基于该文档的解决方案：

URL xsdUrlA = this.getClass().getResource("a.xsd");
URL xsdUrlB = this.getClass().getResource("b.xsd");
URL xsdUrlC = this.getClass().getResource("c.xsd");

SchemaFactory schemaFactory = schemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
//---
String W3C_XSD_TOP_ELEMENT =
"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"yes\"?>\n"
   + "<xs:schema xmlns:xs=\"http://www.w3.org/2001/XMLSchema\" elementFormDefault=\"qualified\">\n"
   + "<xs:include schemaLocation=\"" +xsdUrlA.getPath() +"\"/>\n"
   + "<xs:include schemaLocation=\"" +xsdUrlB.getPath() +"\"/>\n"
   + "<xs:include schemaLocation=\"" +xsdUrlC.getPath() +"\"/>\n"
   +"</xs:schema>";
Schema schema = schemaFactory.newSchema(new StreamSource(new StringReader(W3C_XSD_TOP_ELEMENT), "xsdTop"));

来自 xerces 文档：http://xerces.apache.org/xerces2-j/faq-xs.html

import javax.xml.transform.Source;
import javax.xml.transform.stream.StreamSource;
import javax.xml.validation.Schema;
import javax.xml.validation.SchemaFactory;
import javax.xml.validation.Validator;

...

StreamSource[] schemaDocuments = /* created by your application */;
Source instanceDocument = /* created by your application */;

SchemaFactory sf = SchemaFactory.newInstance(
    "http://www.w3.org/XML/XMLSchema/v1.1");
Schema s = sf.newSchema(schemaDocuments);
Validator v = s.newValidator();
v.validate(instanceDocument);

我最终使用了这个：

import org.apache.xerces.parsers.SAXParser;
import org.xml.sax.SAXException;
import org.xml.sax.SAXParseException;
import org.xml.sax.helpers.DefaultHandler;
import java.io.IOException;
 .
 .
 .
 try {
        SAXParser parser = new SAXParser();
        parser.setFeature("http://xml.org/sax/features/validation", true);
        parser.setFeature("http://apache.org/xml/features/validation/schema", true);
        parser.setFeature("http://apache.org/xml/features/validation/schema-full-checking", true);
        parser.setProperty("http://apache.org/xml/properties/schema/external-noNamespaceSchemaLocation", "http://your_url_schema_location");

        Validator handler = new Validator();
        parser.setErrorHandler(handler);
        parser.parse("file:///" + "/home/user/myfile.xml");

 } catch (SAXException e) {
    e.printStackTrace();
 } catch (IOException ex) {
    e.printStackTrace();
 }


class Validator extends DefaultHandler {
    public boolean validationError = false;
    public SAXParseException saxParseException = null;

    public void error(SAXParseException exception)
            throws SAXException {
        validationError = true;
        saxParseException = exception;
    }

    public void fatalError(SAXParseException exception)
            throws SAXException {
        validationError = true;
        saxParseException = exception;
    }

    public void warning(SAXParseException exception)
            throws SAXException {
    }
}

记得更改：

1） xsd 文件位置的参数“http：//your_url_schema_location”。

2） 指向 xml 文件的字符串“/home/user/myfile.xml”。

我不必设置变量：-Djavax.xml.validation.SchemaFactory:http://www.w3.org/2001/XMLSchema=org.apache.xerces.jaxp.validation.XMLSchemaFactory

我遇到了同样的问题，经过调查找到了这个解决方案。它对我有用。

Enum要设置不同的：XSDs

public enum XsdFile {
    // @formatter:off
    A("a.xsd"),
    B("b.xsd"),
    C("c.xsd");
    // @formatter:on

    private final String value;

    private XsdFile(String value) {
        this.value = value;
    }

    public String getValue() {
        return this.value;
    }
}

验证方法：

public static void validateXmlAgainstManyXsds() {
    final SchemaFactory schemaFactory = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);

    String xmlFile;
    xmlFile = "example.xml";

    // Use of Enum class in order to get the different XSDs
    Source[] sources = new Source[XsdFile.class.getEnumConstants().length];
    for (XsdFile xsdFile : XsdFile.class.getEnumConstants()) {
        sources[xsdFile.ordinal()] = new StreamSource(xsdFile.getValue());
    }

    try {
        final Schema schema = schemaFactory.newSchema(sources);
        final Validator validator = schema.newValidator();
        System.out.println("Validating " + xmlFile + " against XSDs " + Arrays.toString(sources));
        validator.validate(new StreamSource(new File(xmlFile)));
    } catch (Exception exception) {
        System.out.println("ERROR: Unable to validate " + xmlFile + " against XSDs " + Arrays.toString(sources)
                + " - " + exception);
    }
    System.out.println("Validation process completed.");
}

为了以防万一，仍然有人来这里寻找针对多个 XSD 验证 xml 或对象的解决方案，我在这里提到它

//Using **URL** is the most important here. With URL, the relative paths are resolved for include, import inside the xsd file. Just get the parent level xsd here (not all included xsds).

URL xsdUrl = getClass().getClassLoader().getResource("my/parent/schema.xsd");

SchemaFactory schemaFactory = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
Schema schema = schemaFactory.newSchema(xsdUrl);

JAXBContext jaxbContext = JAXBContext.newInstance(MyClass.class);
Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();
unmarshaller.setSchema(schema);

/* If you need to validate object against xsd, uncomment this
ObjectFactory objectFactory = new ObjectFactory();
JAXBElement<MyClass> wrappedObject = objectFactory.createMyClassObject(myClassObject); 
marshaller.marshal(wrappedShipmentMessage, new DefaultHandler());
*/

unmarshaller.unmarshal(getClass().getClassLoader().getResource("your/xml/file.xml"));

如果所有 XSD 都属于同一命名空间，则创建一个新的 XSD 并将其他 XSD 导入其中。然后在 java 中使用新的 XSD 创建架构。

Schema schema = xmlSchemaFactory.newSchema(
    new StreamSource(this.getClass().getResourceAsStream("/path/to/all_in_one.xsd"));

all_in_one.xsd：

<?xml version="1.0" encoding="UTF-8"?>
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema"
 xmlns:ex="http://example.org/schema/" 
 targetNamespace="http://example.org/schema/" 
 elementFormDefault="unqualified"
 attributeFormDefault="unqualified">

    <xs:include schemaLocation="relative/path/to/a.xsd"></xs:include>
    <xs:include schemaLocation="relative/path/to/b.xsd"></xs:include>
    <xs:include schemaLocation="relative/path/to/c.xsd"></xs:include>

</xs:schema>