如何在Java中检查XSD是否有效

如何检查给定文件是否为Java 7中的有效XSD文件（没有internet连接）

这不是重复的。我不想对照XSD检查XML，而是检查XSD本身是否有效

到目前为止，我所尝试的：

@Slf4j
public class Program {

  /**
   * Sample main method.
   * 
   * @param args
   *          program arguments
   */
  public static void main(String[] args) {
    try {
      log.info("Program has started.");
      DocumentBuilder parser = DocumentBuilderFactory.newInstance()
          .newDocumentBuilder();
      Document document = parser.parse(new File("test.xsd"));
      SchemaFactory factory = SchemaFactory
          .newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
      Schema schema = factory.newSchema(new URL(
          "http://www.w3.org/2001/XMLSchema"));
...

      log.info("Program has finished - ok.");
    } catch (Exception ex) {
      ex.printStackTrace();
    }
  }
}

问题是：

即使test.xsd有效，也会引发一些奇怪的异常

正在从internet获取验证模式，但我必须在没有 互联网连接

例外情况是：

org.xml.sax.SAXParseException; systemId: http://www.w3.org/2001/XMLSchema; lineNumber: 7; columnNumber: 20; s4s-elt-character: Non-whitespace characters are not allowed in schema elements other than 'xs:appinfo' and 'xs:documentation'. Saw 'XML Schema'.
at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.createSAXParseException(Unknown Source)
at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.error(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLErrorReporter.reportError(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLErrorReporter.reportError(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.xs.opti.SchemaDOMParser.characters(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.scanDocument(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.xs.opti.SchemaParsingConfig.parse(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.xs.opti.SchemaParsingConfig.parse(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.xs.opti.SchemaDOMParser.parse(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.xs.traversers.XSDHandler.getSchemaDocument(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.xs.traversers.XSDHandler.parseSchema(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.xs.XMLSchemaLoader.loadSchema(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.xs.XMLSchemaLoader.loadGrammar(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.xs.XMLSchemaLoader.loadGrammar(Unknown Source)
at com.sun.org.apache.xerces.internal.jaxp.validation.XMLSchemaFactory.newSchema(Unknown Source)
at javax.xml.validation.SchemaFactory.newSchema(Unknown Source)
at javax.xml.validation.SchemaFactory.newSchema(Unknown Source)
at o2.xml.core.Program.main(Program.java:39)

问题可能是在指定模式时，所以我应该指定什么来检查XSD？是否有其他预构建常量或什么？

这一个（基于）对我来说似乎没问题：

URL schemaFile = new URL("https://www.w3.org/2001/XMLSchema.xsd");
Source xmlFile = new StreamSource(XMLSchemaTest.class.
getResourceAsStream("mySchema.xsd"));
SchemaFactory schemaFactory =   
     SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
try {
     Schema schema = schemaFactory.newSchema(schemaFile);
     Validator validator = schema.newValidator();
     validator.validate(xmlFile);
     System.out.println("is valid");
} catch (SAXException e) {
     System.out.println(xmlFile.getSystemId() + " is NOT valid reason:" + e);
} catch (IOException e) {
     e.printStackTrace();
}

---

我也尝试过使用，但这一次在主模式读取过程中产生错误

可能重复否，我不想根据XSD检查XML，但要检查XSD的有效性。@TobySpeight ok，updated@TobySpeight增加了例外情况，更详细地说明了问题可能是什么是XMLSchemaTest？编辑：我认为这是该代码所在的类的名称。