JAXB 使用包含内容解码 xml

Question

我下面有一个名为“TheFile.xml”的 xml

    <!DOCTYPE page [
      <!ENTITY class_common SYSTEM "class_common.xml" >]>

    <api xmlns:denp="http://intranet.denali.com/wiki/DocBook_profiling">

    &class_common;

正如您所看到的，上面的行引用了另一个名为“class_common.xml”的 xml 文件。然而，当我这样做时

    cl = class_xml.ObjectFactory.class.getClassLoader();
    context = JAXBContext.newInstance("class_xml", cl);
    Unmarshaller u = context.createUnmarshaller();
    api = (Api) u.unmarshal(new FileInputStream(classXMLFile));

我收到一些如下错误。我错过了什么吗？谢谢

线程“main”中出现异常 javax.xml.bind.UnmarshalException - 带有链接异常:

[org.xml.sax.SAXParseException; lineNumber: 8; columnNumber: 17; External Entity: Failed to read external document 'class_common.xml', because 'file' access is not allowed due to restriction set by the accessExternalDTD property.]

Answer 1

好的，这就是答案，Java 代码需要如下

    cl = class_xml.ObjectFactory.class.getClassLoader();
    context = JAXBContext.newInstance("class_xml", cl);
    Unmarshaller u = context.createUnmarshaller();

    SAXParserFactory spf = SAXParserFactory.newInstance();

    spf.setXIncludeAware(true);
    spf.setNamespaceAware(true);
    spf.setValidating(true); // Not required for JAXB/XInclude

    XMLReader xr = (XMLReader) spf.newSAXParser().getXMLReader();
    SAXSource source = new SAXSource(xr, new InputSource(new   
        FileInputStream(classXMLFile)));

    api = (Api) u.unmarshal(source);