17

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
        }
    }

}
4

4 回答 4

16

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

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

对应的Javadoc

于 2012-12-10T19:36:18.640 回答
8

您已经编组到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();
于 2012-12-09T15:09:33.690 回答
2

Marshaller#marshall(...)方法将 OutputStream 或 Writer 作为参数。如果你看过的话，你肯定会在 API 中找到这个。

于 2012-12-09T15:09:51.800 回答
2

这只是从 java 对象到 xml 文件的转换过程。它与序列化非常相似，您必须确定序列化和编组。在这里，我已经完成了编组的示例。您可以以类似的方式进行解组。

带有 jaxp 注释的 bean 类：

package com.ofs.swinapps;

import javax.xml.bind.annotation.XmlRootElement;

    @XmlRootElement
    public class Employee {
    private String name;
    private String id;
    private String department;
    private int age;
    private int salary;

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
    public String getId() {
        return id;
    }
    public void setId(String id) {
        this.id = id;
    }

    public String getDepartment() {
        return department;
    }

    public void setDepartment(String department) {
        this.department = department;
    }

    public int getAge() {
        return age;
}

    public void setAge(int age) {
        this.age = age;
    }

    public int getSalary() {
        return salary;
    }
    public void setSalary(int salary) {
        this.salary = salary;
    }
    }

编组：

import java.io.File;
import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Marshaller;

public class Java2XMLExample {
    public static void main(String[] args) throws JAXBException {
 Employee employee = new Employee();
 employee.setName("Kowthal ganesh");
 employee.setAge(23);
 employee.setDepartment("Chola-ccms");
 employee.setId("947");
 employee.setSalary(8333);
 File file = new File("D:/build.xml");
 JAXBContext jaxbContext = JAXBContext.newInstance(Employee.class);
 Marshaller jaxbMarshaller = jaxbContext.createMarshaller();
 jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
 jaxbMarshaller.marshal(employee, file);
    }
}
于 2013-11-27T09:25:08.177 回答