JAXB编组Java以输出XML文件

Question

问题是如何生成XML文件输出而不是system.out？

package jaxbintroduction;

import java.io.FileOutputStream;
import java.io.OutputStream;

public class Main {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
        itemorder.Book quickXML = new itemorder.Book();

        quickXML.setAuthor("Sillyme");
        quickXML.setDescription("Dummie book");
        quickXML.setISBN(123456789);
        quickXML.setPrice((float)12.6);
        quickXML.setPublisher("Progress");
        quickXML.setTitle("Hello World JAVA");

        try {            
            javax.xml.bind.JAXBContext jaxbCtx = javax.xml.bind.JAXBContext.newInstance(quickXML.getClass().getPackage().getName());
            javax.xml.bind.Marshaller marshaller = jaxbCtx.createMarshaller();
            marshaller.setProperty(javax.xml.bind.Marshaller.JAXB_ENCODING, "UTF-8"); //NOI18N
            marshaller.setProperty(javax.xml.bind.Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE);
            marshaller.marshal(quickXML, System.out);
            OutputStream os = new FileOutputStream( "nosferatu.xml" );
            marshaller.marshal( quickXML, os );

        } catch (javax.xml.bind.JAXBException ex) {
            // XXXTODO Handle exception
            java.util.logging.Logger.getLogger("global").log(java.util.logging.Level.SEVERE, null, ex); //NOI18N
        }
    }

}

Answer 1

如果您使用的是JAXB 2.1或更高版本,则可以直接编组java.io.File对象:

 File file = new File( "nosferatu.xml" );
 marshaller.marshal( quickXML, file );

相应的Javadoc

• http://docs.oracle.com/javase/6/docs/api/javax/xml/bind/Marshaller.html#marshal%28java.lang.Object,%20java.io.File%29

Answer 2

你已经在编组了nosferatu.xml.只需删除或评论该行:

marshaller.marshal(quickXML, System.out);

如果您不想显示输出并关闭OutputStream:

os.close();