JAXB 实现java对象与xml之间互相转换(二)

墨蓝 2022-06-13 06:16 211阅读 0赞

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> {
    
    	@Override
    	public Address unmarshal(String v) throws Exception {
    		AddressImpl address = new AddressImpl();
    		address.setAddress(v);
    		return address;
    	}
    
    	@Override
    	public 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>