java 没有 XmlRootElement 注释的 JAXB 解组？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/33823139/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
JAXB unmarshalling without XmlRootElement annotation?
提问by Kaleb
Is there any way we can un-marshall for a class without @XmlRootElement annotation? Or are we obligated to enter the annotation?
有什么方法可以取消编组没有@XmlRootElement 注释的类？还是我们有义务输入注释？
for example:
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;
}
}
and let the unmarshalling code for properly annotated class be like:
并让正确注释的类的解组代码如下：
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();
}
leaving out the details.
省略细节。
回答by Xstian
Following code is used to marshall and unmarshall withot @XmlRootElement
以下代码用于编组和解组没有 @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();
}
}
Above code works only if you adding @XmlAccessorType(XmlAccessType.PROPERTY)on Customer class, or make private all attributes.
仅当您添加@XmlAccessorType(XmlAccessType.PROPERTY)Customer 类或将所有属性设为私有时，以上代码才有效。
回答by Abhishek K
If you cannot add XmlRootElement to existing bean you can also create a holder class and mark it with annotation as XmlRootElement. Example below:-
如果您无法将 XmlRootElement 添加到现有 bean，您还可以创建一个持有者类并使用注释将其标记为 XmlRootElement。下面的例子：-
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;
}
}
