7

我有以下 XML：

<ns2:Person name="John" age="20" />

我想将它解组为从 XSD 生成的 JAXB 对象 Person。

这是我正在运行的代码：

JAXBContext context = JAXBContext.newInstance(PersoEntity.class);
Unmarshaller um = context.createUnmarshaller();
StringReader sr = new StringReader(xml);
Person p = (Person)um.unmarshal(sr);

令人惊讶的是，我得到以下异常：

javax.xml.bind.UnmarshalException
 - with linked exception:
[org.xml.sax.SAXParseException: The prefix "ns2" for element "ns2:Person" is not bound.]

我该如何解决？谢谢

4

2 回答 2

6

获取片段

您当前获取 XML 片段的方式导致名称空间声明丢失。在您的片段ns2中不再是前缀，您只有一个带有冒号的元素名称（ns2:Person）。这将导致命名空间感知解析器出现问题。下面的文章可能是您获取 XML 片段的更好方法：

处理您的用例

使用您拥有的 XML 片段，您可以创建一个XMLFilter从 XML 元素中删除前缀的片段，然后利用 JAXBUnmarshallerHandler进行解组。

演示

package forum11968399;

import java.io.StringReader;
import javax.xml.bind.*;
import javax.xml.parsers.*;
import org.xml.sax.*;
import org.xml.sax.helpers.XMLFilterImpl;

public class Demo {

    private static final String xml = "<ns2:Person name='John' age='20' />";

    public static void main(String[] args) throws Exception {
        SAXParserFactory spf = SAXParserFactory.newInstance();
        SAXParser sp = spf.newSAXParser();
        XMLReader xmlReader = sp.getXMLReader();
        XMLFilter xmlFilter = new MyXMLFilter(xmlReader);

        JAXBContext context = JAXBContext.newInstance(PersonEntity.class);
        Unmarshaller um = context.createUnmarshaller();
        UnmarshallerHandler unmarshallerHandler = um.getUnmarshallerHandler();
        xmlFilter.setContentHandler(unmarshallerHandler);

        StringReader sr = new StringReader(xml);
        xmlFilter.parse(new InputSource(sr));
        PersonEntity p = (PersonEntity) unmarshallerHandler.getResult();
    }

    private static class MyXMLFilter extends XMLFilterImpl {

        public MyXMLFilter(XMLReader xmlReader) {
            super(xmlReader);
        }

        @Override
        public void startElement(String uri, String localName, String qName,
                Attributes attributes) throws SAXException {
            int colonIndex = qName.indexOf(':');
            if(colonIndex >= 0) {
                qName = qName.substring(colonIndex + 1);
            }
uri = XML_NAMESPACE; //to prevent unknown XML element exception, we have to specify the namespace here
            super.startElement(uri, localName, qName, attributes);
        }

        @Override
        public void endElement(String uri, String localName, String qName)
                throws SAXException {
            int colonIndex = qName.indexOf(':');
            if(colonIndex >= 0) {
                qName = qName.substring(colonIndex + 1);
            }
            super.endElement(uri, localName, qName);
        }

    }

}

个人实体

package forum11968399;

import javax.xml.bind.annotation.*;

@XmlRootElement(name="Person")
@XmlAccessorType(XmlAccessType.FIELD)
public class PersonEntity {

    @XmlAttribute
    private String name;

    @XmlAttribute
    private int age;

}
于 2012-08-15T13:58:03.077 回答
1

您最好的选择可能是将所需元素嵌套在绑定命名空间的另一个元素中。将它绑定到什么并不重要，只需将其设为可解析的有效 XML 文档即可。然后你可以按声明的类型解组

JAXBContext context = JAXBContext.newInstance(Person.class);
Unmarshaller um = context.createUnmarshaller();
String xml = "<ns2:Person name=\"John\" age=\"20\" />";
String xmlWithPrefixMapped = "<ns2:FakeElement xmlns:ns2=\"someuri\">" + xml + "</ns2:FakeElement>";
StringReader sr = new StringReader(xmlWithPrefixMapped);
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
dbf.setNamespaceAware(true);
DocumentBuilder db = dbf.newDocumentBuilder();
Document doc = db.parse(new InputSource(sr));
Node n = (Node) doc.getDocumentElement().getFirstChild();
JAXBElement<Person> personElement = um.unmarshal(n, Person.class);
Person p = personElement.getValue();
于 2012-08-15T11:36:36.153 回答