没有 XmlRootElement 注释的 JAXB 解组?
JAXB unmarshalling without XmlRootElement annotation?
有没有什么方法可以在没有 @XmlRootElement 注释的情况下为 class 解组?还是我们有义务输入注释?
例如:
public class Customer {
private String name;
private int age;
private int id;
public String getName() {
return name;
}
@XmlElement
public void setName(String name) {
this.name = name;
}
public int getAge() {
return age;
}
@XmlElement
public void setAge(int age) {
this.age = age;
}
public int getId() {
return id;
}
@XmlAttribute
public void setId(int id) {
this.id = id;
}
}
并让正确注释的 class 的解组代码如下所示:
try {
File file = new File("C:\file.xml");
JAXBContext jaxbContext = JAXBContext.newInstance(Customer.class);
Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
Customer customer = (Customer) jaxbUnmarshaller.unmarshal(file);
System.out.println(customer);
} catch (JAXBException e) {
e.printStackTrace();
}
省略细节。
以下代码用于编组和解组而不用 @XmlRootElement
public static void main(String[] args) {
try {
StringWriter stringWriter = new StringWriter();
Customer c = new Customer();
c.setAge(1);
c.setName("name");
JAXBContext jaxbContext = JAXBContext.newInstance(Customer.class);
Marshaller marshaller = jaxbContext.createMarshaller();
marshaller.marshal(new JAXBElement<Customer>( new QName("", "Customer"), Customer.class, null, c), stringWriter);
Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
InputStream is = new ByteArrayInputStream(stringWriter.toString().getBytes());
JAXBElement<Customer> customer = (JAXBElement<Customer>) jaxbUnmarshaller.unmarshal(new StreamSource(is),Customer.class);
c = customer.getValue();
} catch (JAXBException e) {
e.printStackTrace();
}
}
只有在客户 class 上添加 @XmlAccessorType(XmlAccessType.PROPERTY)
或将所有属性设为私有时,以上代码才有效。
如果您无法将 XmlRootElement 添加到现有 bean,您还可以创建一个持有者 class 并使用注释将其标记为 XmlRootElement。以下示例:-
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlRootElement;
@XmlRootElement
@XmlAccessorType(XmlAccessType.FIELD)
public class CustomerHolder
{
private Customer cusotmer;
public Customer getCusotmer() {
return cusotmer;
}
public void setCusotmer(Customer cusotmer) {
this.cusotmer = cusotmer;
}
}
有没有什么方法可以在没有 @XmlRootElement 注释的情况下为 class 解组?还是我们有义务输入注释?
例如:
public class Customer {
private String name;
private int age;
private int id;
public String getName() {
return name;
}
@XmlElement
public void setName(String name) {
this.name = name;
}
public int getAge() {
return age;
}
@XmlElement
public void setAge(int age) {
this.age = age;
}
public int getId() {
return id;
}
@XmlAttribute
public void setId(int id) {
this.id = id;
}
}
并让正确注释的 class 的解组代码如下所示:
try {
File file = new File("C:\file.xml");
JAXBContext jaxbContext = JAXBContext.newInstance(Customer.class);
Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
Customer customer = (Customer) jaxbUnmarshaller.unmarshal(file);
System.out.println(customer);
} catch (JAXBException e) {
e.printStackTrace();
}
省略细节。
以下代码用于编组和解组而不用 @XmlRootElement
public static void main(String[] args) {
try {
StringWriter stringWriter = new StringWriter();
Customer c = new Customer();
c.setAge(1);
c.setName("name");
JAXBContext jaxbContext = JAXBContext.newInstance(Customer.class);
Marshaller marshaller = jaxbContext.createMarshaller();
marshaller.marshal(new JAXBElement<Customer>( new QName("", "Customer"), Customer.class, null, c), stringWriter);
Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
InputStream is = new ByteArrayInputStream(stringWriter.toString().getBytes());
JAXBElement<Customer> customer = (JAXBElement<Customer>) jaxbUnmarshaller.unmarshal(new StreamSource(is),Customer.class);
c = customer.getValue();
} catch (JAXBException e) {
e.printStackTrace();
}
}
只有在客户 class 上添加 @XmlAccessorType(XmlAccessType.PROPERTY)
或将所有属性设为私有时,以上代码才有效。
如果您无法将 XmlRootElement 添加到现有 bean,您还可以创建一个持有者 class 并使用注释将其标记为 XmlRootElement。以下示例:-
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlRootElement;
@XmlRootElement
@XmlAccessorType(XmlAccessType.FIELD)
public class CustomerHolder
{
private Customer cusotmer;
public Customer getCusotmer() {
return cusotmer;
}
public void setCusotmer(Customer cusotmer) {
this.cusotmer = cusotmer;
}
}