Java 使用没有 namespace 的 XSD 验证 XML

Question

我正在尝试验证以下 XML

<query>
 <colors logic="AND">
  <color main="BLUE" tone="DARK" operator="=" />
 </colors>
</query>

使用以下 XSD

<?xml version="1.0" encoding="UTF-8"?>
<xsd:schema xmlns:xsd="http://www.w3.org/2001/XMLSchema">
 <xsd:complexType name="color">
     <xsd:attribute name="main" type="xsd:string" use="required"/>
     <xsd:attribute name="tone" type="xsd:string" use="required"/>
     <xsd:attribute name="operator" type="xsd:string"/>
 </xsd:complexType>
 <xsd:complexType name="colors">
     <xsd:sequence>
         <xsd:element name="color" type="color" maxOccurs="unbounded">   </xsd:element>
     </xsd:sequence>
     <xsd:attribute name="logic" type="xsd:string" use="required"/>
 </xsd:complexType>
 <xsd:complexType name="query">
     <xsd:sequence>
         <xsd:element name="colors" type="colors" maxOccurs="2"></xsd:element>
     </xsd:sequence>
 </xsd:complexType>
 <xsd:element name="query" type="query"></xsd:element>
</xsd:schema>

所以...我想在没有任何 namespace 的情况下验证 XML。我无法更改 XML，因为它是由另一个应用程序生成的，我只想在服务器端保证客户端正在发送正确的请求。

当我尝试根据 XSD 验证 XML 时，我收到以下异常消息:

cvc-elt.1: Cannot find the declaration of element 'query'

我已经搜索并找到了类似 this 的解决方案和 this但没有成功

解决方案(感谢@Traroth 为我指明了正确的方向)---

这是我验证它的方式:

我有这个功能:

public static Document buildValidRequest(String content, Schema xsd) throws SAXParseException, SAXException, IOException, ParserConfigurationException {

    DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();

    factory.setValidating(false);
    factory.setNamespaceAware(true);

    factory.setSchema(xsd);

    XMLSimpleErrorHandler errorHandler = new XMLSimpleErrorHandler();

    DocumentBuilder builder = factory.newDocumentBuilder();
    builder.setErrorHandler(errorHandler);

    StringReader reader = new StringReader(content);

    Document validXML = builder.parse(new InputSource(reader));

    if (errorHandler.getException() != null) {
        throw errorHandler.getException();
    }

    return validXML;

}

这个类来处理错误:

public class XMLSimpleErrorHandler implements ErrorHandler {

private SAXParseException exception;

@Override
public void warning(SAXParseException e) {
    this.exception = e;
}

@Override
public void error(SAXParseException e) {
    this.exception = e;
}

@Override
public void fatalError(SAXParseException e) {
    this.exception = e;
}

public SAXParseException getException() {
    return exception;
}
}

以及获取架构的方法:

private static Schema getSchema(String xsdPath) throws SAXException, IOException {
    InputStream resourceAsStream = null;
    try {
        ServiceManager.getInstance().getLoggerManager().debug(RESTInitServlet.LOGCONTEXT, TAG, "Retrieving schema: "+xsdPath);
        resourceAsStream = getInstance().getClass().getResourceAsStream(xsdPath);
        SchemaFactory schemaFactory = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
        Source schemaSource = new StreamSource(resourceAsStream);
        Schema schema = schemaFactory.newSchema(schemaSource);
        return schema;
    } finally {
        if (resourceAsStream != null) {
            resourceAsStream.close();
        }
    }

}

没关系:最奇怪的是:适用于运行在 Windows 7 上的 Tomcat 6；在 Linux 上运行的 jboss 中不工作...

Answer 1

您应该在这里找到答案:http://www.edankert.com/validate.html#Validate_using_external_Schema