如何在JAXB中解组xml时处理特殊字符

Question

my test xml content

    <p id="033" num="03">geopotent change&#x2 high.</p>

And I run Jaxb Unmarshalling, but I'm getting exception

    09:58:43.748 ERROR [main][net.ServiceImpl] Parsing Error: 
    javax.xml.bind.UnmarshalException- with linked exception:    
    [javax.xml.stream.XMLStreamException: ParseError at [row,col]:
    [161,306]Message: String "&#]

my Jaxb unmarshal source is

    JAXBContext jaxbContext = JAXBContext.newInstance(CnDocument.class);
    Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
    document = (CnDocument) jaxbUnmarshaller.unmarshal(xmlFile);

How can I escape those characters? (&#x2)

Answer 1

You get a JAXB parse error because your XML content is not well-formed. It should be  (with semicolon), not &#x2 .

Answer 2

<p id="033" num="03">geopotent change&#x2 high.</p> is not valid XML. Either "escape" the ampersand character using & or make the character reference complete by adding a semicolon:  .

Answer 3

This is one way you can do:

import java.io.IOException;
import java.io.StringWriter;
import java.io.Writer;

import com.sun.xml.bind.marshaller.CharacterEscapeHandler;

public class XmlCharacterHandler implements CharacterEscapeHandler {

    public void escape(char[] buf, int start, int len, boolean isAttValue,
            Writer out) throws IOException {
        StringWriter buffer = new StringWriter();

        for (int i = start; i < start + len; i++) {
            buffer.write(buf[i]);
        }

        String st = buffer.toString();

        if (!st.contains("CDATA")) {
            st = buffer.toString().replace("&", "&amp;").replace("<", "&lt;")
                .replace(">", "&gt;").replace("'", "&apos;")
                .replace("\"", "&quot;");

        }
        out.write(st);
        System.out.println(st);
    }

}

While marshaling :

marshaller.setProperty(CharacterEscapeHandler.class.getName(),
                new XmlCharacterHandler());