Question

解组XML文件后，XML文件中属性的所有属性的值都为NULL（fileDateTime，fileId等等）

我真的不明白为什么我有正确的注释@XmlAttribute(name = "FileDateTime")和@XmlAttribute(name = "FileId")

正如您所看到的，我不使用任何命名空间（因此我认为不是命名空间问题！）

我正在使用JDK 1.6，Sax 2.0.1和XercesImpl 2.9.1

感谢您的帮助。

的test.xml

<KeyImport_file FileDateTime="2013-05-30T09:00:00" FileId="KeyImport_source_20121231124500">
    <!--1 or more repetitions:-->
    <record record_number="10">
    ...
    </record>
</KeyImport_file>

KeyImportFile.java

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "", propOrder = {
    "record"
})
@XmlRootElement(name = "KeyImport_file")
public class KeyImportFile {

    @XmlElement(required = true)
    protected List<KeyImportFile.Record> record;

    @XmlAttribute(name = "FileDateTime")
    @XmlSchemaType(name = "dateTime")
    protected XMLGregorianCalendar fileDateTime;

    @XmlAttribute(name = "FileId")
    protected String fileId;
etc...
etc...

解析方法（unmarshal＆amp; XSD验证）：

import org.xml.sax.InputSource;
import org.xml.sax.XMLReader;

import javax.xml.XMLConstants;
import javax.xml.bind.JAXBContext;
import javax.xml.bind.UnmarshallerHandler;
import javax.xml.parsers.SAXParser;
import javax.xml.parsers.SAXParserFactory;
import javax.xml.transform.stream.StreamSource;
import javax.xml.validation.Schema;
import javax.xml.validation.SchemaFactory;
import java.io.InputStream;

private KeyImportFile parseXML(final InputStream xmlInputStream, final StreamSource xsdSource)
        throws Exception
{
    KeyImportFile keyImportFile;

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

    JAXBContext jc = JAXBContext.newInstance(KeyImportFile.class);

    UnmarshallerHandler unmarshallerHandler = jc.createUnmarshaller().getUnmarshallerHandler();

    SAXParserFactory saxParserFactory = SAXParserFactory.newInstance();
    saxParserFactory.setSchema(schema);

    SAXParser saxParser = saxParserFactory.newSAXParser();
    XMLReader xmlReader = saxParser.getXMLReader();
    xmlReader.setContentHandler(unmarshallerHandler);
    xmlReader.setErrorHandler(keyImportErrorHandler);

    InputSource inputSource = new InputSource(xmlInputStream);
    xmlReader.parse(inputSource);
    xmlInputStream.close();

    keyImportFile = (KeyImportFile) unmarshallerHandler.getResult();

    return keyImportFile;
}

编辑

只需更改我的解析方法而不使用sax就可以了。知道为什么吗？我想使用sax和jabx来解决性能问题。

KeyImportFile keyImportFile;

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

JAXBContext jc = JAXBContext.newInstance(KeyImportFile.class);

Unmarshaller unmarshaller = jc.createUnmarshaller();
unmarshaller.setSchema(schema);
keyImportFile = (KeyImportFile) unmarshaller.unmarshal(xmlInputStream);

Answer 1

<强>解决

似乎JAXB无法正常使用SAX解析器，除非将解析器设置为名称空间感知。

刚添加此行并且工作正常

saxParserFactory.setNamespaceAware(true);

JAXB unmarshal为属性返回null

1 个答案: