Question

Java XML Schema验证 - ResourceResolver麻烦

我正在尝试将一个组件构建到一个Web应用程序中，该应用程序将根据一组模式验证不同的XML文档。

我在java包com.example.xml中有验证器类，然后我有一个＆＃34;包＆＃34;对于schemas com.example.xml.Schemas，它们根据命名空间和任何包含的模式进行组织，如下所示：

com.example.xml/SchemaValidator.java

com.example.xml.Schemas/2008/07/05/Message.xsd
com.example.xml.Schemas/2008/07/05/Person.xsd
com.example.xml.Schemas/2008/07/05/Address.xsd

com.example.xml.Schemas/2010/09/21/Message.xsd
com.example.xml.Schemas/2010/09/21/Organization.xsd
com.example.xml.Schemas/2010/09/21/Person.xsd
com.example.xml.Schemas/2010/09/21/Address.xsd

因此，大多数文档元素都称为Message，但在不同的名称空间和不同的内部。通过映射，我可以确定正确的模式所在的位置，然后我加载与根元素名称匹配的XSD - 但为了使解析器能够读取包含的模式，我尝试了一个org.w3c.dom.ls .LSResourceResolver实现基于systemId获取正确的文件。

public class ResourceResolver implements LSResourceResolver {
    private String basePath;

    public ResourceResolver( String baseDirectory ){
        basePath = baseDirectory;
        if( !basePath.endsWith("/")){
            basePath += "/";
        }
    }

    @Override
    public LSInput resolveResource(String type, String namespaceURI, String publicId, String systemId, String baseURI) {
        System.out.println("Resolving: " + type + ", " + namespaceURI + ", " +  publicId  + ", " + systemId + ", " + baseURI + " (basepath:" + basePath + ")");
        String mypath = basePath + systemId;
        InputStream resourceAsStream = getClass().getResourceAsStream(mypath);
        if( resourceAsStream == null){ System.out.println("Ups! Stream is null"); }

        String contents = null;
        byte[] bytes = null;
        try {
            bytes = new byte[resourceAsStream.available()];
            resourceAsStream.read(bytes);
            contents = new String(bytes, Charset.forName("UTF-8"));
        } catch (IOException e) { } 
        finally{
            try { resourceAsStream.close(); } 
            catch (IOException e) { }
        }
        Input xmlInput = new Input(resourceAsStream, type, namespaceURI, publicId, systemId, basePath);    
        xmlInput.setStringData(contents);  // avoid problems with inputstream position
        return xmlInput;
    }
}

我的ResourceResolver以这种方式使用：

public boolean validate(Document doc, String xsdPath) throws SAXException, IOException, ParserConfigurationException {
    SchemaFactory schemaFactory = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
    String rootElmtName = doc.getDocumentElement().getLocalName();
    String resourcePath = "/com/example/xml/" + xsdPath ;
    InputStream topSchema = getClass().getResourceAsStream(xsdPath + "/" + rootElmtName + ".xsd");
    Source schemaSource = new StreamSource(topSchema);
    ResourceResolver resolver = new ResourceResolver(resourcePath);
    schemaFactory.setResourceResolver(resolver);
    Schema schema = schemaFactory.newSchema(schemaSource);
    Validator validator = schema.newValidator();
    validator.setResourceResolver(resolver);
    Source source = new DOMSource(doc);

    ValidationErrorHandler handler = new ValidationErrorHandler();
    validator.setErrorHandler(handler);
    validator.validate(source);
    if(handler.hasErrors()){
        for( ValidationError e : handler.getErrors()){
            log(e);
        }
        return false;
    }
    return true;
}

然而，它不起作用！它解析了一些架构资源然后停止我尝试在我的资源解析器中在Eclipse中设置断点，并且输入流不为空 - 它指向包含XSD的正确文件。因此，Java XML解析器类中的某些内容必须从我的解析器中获得一些其他行为。

Resolving: http://www.w3.org/2001/XMLSchema, http://com/example/xml/schemas/2005/08/07/, null, Letters.xsd, file:///com/example/xml/Schemas/2005/08/07/Message.xsd (basepath:/com/example/xml/Schemas/2005/08/07/)
Resolving: http://www.w3.org/2001/XMLSchema, http://com/example/xml/schemas/2005/08/07/, null, GeneralElements.xsd, file:///com/example/xml/Schemas/2005/08/07/Message.xsd (basepath:/com/example/xml/Schemas/2005/08/07/)
Resolving: http://www.w3.org/2001/XMLSchema, http://com/example/xml/schemas/2005/08/07/, null, GeneralTypes.xsd, file:///com/example/xml/Schemas/2005/08/07/Message.xsd (basepath:/com/example/xml/Schemas/2005/08/07/)
Exception in thread "main" org.xml.sax.SAXParseException: src-resolve: Cannot resolve the name 'Letters' to a(n) 'group' component.
    at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.createSAXParseException(ErrorHandlerWrapper.java:195)
    at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.error(ErrorHandlerWrapper.java:131)
    at com.sun.org.apache.xerces.internal.impl.XMLErrorReporter.reportError(XMLErrorReporter.java:384)
    at com.sun.org.apache.xerces.internal.impl.xs.traversers.XSDHandler.reportSchemaErr(XSDHandler.java:2537)
    at com.sun.org.apache.xerces.internal.impl.xs.traversers.XSDHandler.reportSchemaError(XSDHandler.java:2528)
    at com.sun.org.apache.xerces.internal.impl.xs.traversers.XSDHandler.getGlobalDecl(XSDHandler.java:1472)
    at com.sun.org.apache.xerces.internal.impl.xs.traversers.XSDGroupTraverser.traverseLocal(XSDGroupTraverser.java:72)

Message.xsd的内容包含如下内容：

<xs:include schemaLocation="Letters.xsd"/>
<xs:include schemaLocation="GeneralElements.xsd"/>
<xs:include schemaLocation="GeneralTypes.xsd"/>
<xs:element name="Message">

- 而Letters.xsd也包含

<xs:include schemaLocation="GeneralElements.xsd"/>
<xs:include schemaLocation="GeneralTypes.xsd"/>
<xs:group name="Letters">

大纲属于ResourceResolver，输入类来自另一个SO帖子 - 所以我可能误解了一些东西。

这是我创建资源解析器的正确方法，还是我完全错过了什么？

具有自定义资源解析程序的Java XML Schema验证程序无法解析元素

0 个答案: