如何让JAXB unmarshaller忽略前缀？

Question

我有以下XML：

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

我想将它解组到从XSD生成的JAXB对象的人。

这是代码，我跑：

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，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.] 

如何解决这个问题呢？谢谢

Answer 1

抵达片段

您目前得到的XML片段导致丢失的命名空间声明的方式。在你的片段中ns2不再是一个前缀，你只需要一个带冒号的元素名称（ns2:Person）。这会导致命名空间感知解析器出现问题。下面的文章可能是一个更好的办法，你得到的XML片段：

• http://blog.bdoughan.com/2012/08/handle-middle-of-xml-document-with-jaxb.html

处理您的使用例

使用，你有XML片段，您可以创建一个XMLFilter，它从XML元素中删除前缀，然后利用JAXB的UnmarshallerHandler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); 
     } 

    } 

} 

PersonEntity

package forum11968399; 

import javax.xml.bind.annotation.*; 

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

    @XmlAttribute 
    private String name; 

    @XmlAttribute 
    private int age; 

} 

Answer 2

最好的办法是将所需的元素嵌套到另一个绑定命名空间的元素中。将它绑定到什么并不重要，只要将其制作为有效的XML文档即可解析。然后你可以unmarshal by declared type

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();