JAXB 实现java对象与xml之间互相转换(二)

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前面介绍的对象boy中只包含简单属性，如果boy中还包含其他对象该做如何处理呢？

@XmlRootElement(name="Root")@XmlAccessorType(XmlAccessType.PROPERTY)public class Boy {private String name = "aa";private Address address;public String getName() {return name;}public void setName(String name) {this.name = name;}@XmlJavaTypeAdapter(AddressAdapter.class)public Address getAddress() {return address;}public void setAddress(Address address) {this.address = address;}}

public class Address {private String address;public void setAddress(String address) {this.address = address;}public String getAddress() {return address;}}

此时的boy对象中包含了Address对象，所以此时boy对象中的getAddress()须设置adapter，

创建AddressAdapter类，继承XmlAdapter<ValueType,BoundType>

public class AddressAdapter extends XmlAdapter<String, Address> {@Overridepublic Address unmarshal(String v) throws Exception {AddressImpl address = new AddressImpl();address.setAddress(v);return address;}@Overridepublic String marshal(Address v) throws Exception {return v.getAddress();}}

此时即可进行java对象转换为xml

public class JAXBTest {public static void main(String[] args) throws JAXBException {JAXBContext context = JAXBContext.newInstance(Boy.class);Marshaller marshaller = context.createMarshaller();System.out.println("----------marshaller--------------");Boy boy = new Boy();AddressImpl address = new AddressImpl();address.setAddress("BeiJing");boy.setAddress(address);marshaller.marshal(boy, System.out);}}

最后打印结果为：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?><Root><address>BeiJing</address><name>aa</name></Root>