JAXB和构造函数

Question

我开始学习JAXB,所以我的问题可能非常愚蠢.现在我有类并希望生成XML Schema.遵循这条指令,我得到了例外

IllegalAnnotationExceptions ...没有no-arg默认构造函数.

是啊.我的类没有默认的no-arg构造函数.这太容易了.我有包可见的构造函数/最终方法的类,当然还有参数.我该怎么办 - 创建一些特定的momemto/builder类或者将我的构造函数指定给JAXB(以什么方式？)？谢谢.

Answer 1

JAXB可以使用XML适配器支持这种情况.假设您有以下没有零参数构造函数的对象:

package blog.immutable;

public class Customer {

    private final String name;
    private final Address address;

    public Customer(String name, Address address) {
        this.name = name;
        this.address = address;
    }

    public String getName() {
        return name;
    }

    public Address getAddress() {
        return address;
    }

}

您只需要创建此类的可映射版本:

package blog.immutable.adpater;

import javax.xml.bind.annotation.XmlAttribute;
import blog.immutable.Address;

public class AdaptedCustomer {

    private String name;
    private Address address;

    @XmlAttribute
    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public Address getAddress() {
        return address;
    }

    public void setAddress(Address address) {
        this.address = address;
    }

}

和XML适配器之间进行转换:

package blog.immutable.adpater;

import javax.xml.bind.annotation.adapters.XmlAdapter;
import blog.immutable.Customer;

public class CustomerAdapter extends XmlAdapter<AdaptedCustomer, Customer> {

    @Override
    public Customer unmarshal(AdaptedCustomer adaptedCustomer) throws Exception {
        return new Customer(adaptedCustomer.getName(), adaptedCustomer.getAddress());
    }

    @Override
    public AdaptedCustomer marshal(Customer customer) throws Exception {
        AdaptedCustomer adaptedCustomer = new AdaptedCustomer();
        adaptedCustomer.setName(customer.getName());
        adaptedCustomer.setAddress(customer.getAddress());
        return adaptedCustomer;
    }

}

然后,对于引用Customer类的属性,只需使用@XmlJavaTypeAdapter批注:

package blog.immutable;

import javax.xml.bind.annotation.XmlRootElement;
import javax.xml.bind.annotation.adapters.XmlJavaTypeAdapter;
import blog.immutable.adpater.CustomerAdapter;

@XmlRootElement(name="purchase-order")
public class PurchaseOrder {

    private Customer customer;

    @XmlJavaTypeAdapter(CustomerAdapter.class)
    public Customer getCustomer() {
        return customer;
    }

    public void setCustomer(Customer customer) {
        this.customer = customer;
    }

} 

有关更详细的示例,请参阅:

• http://bdoughan.blogspot.com/2010/12/jaxb-and-immutable-objects.html

Answer 2

您可以使用注释@XmlType并以各种组合使用factoryMethod/factoryClass属性,例如:

@XmlType(factoryMethod="newInstance")
@XmlRootElement
public class PurchaseOrder {
    @XmlElement
    private final String address;
    @XmlElement
    private final Customer customer;

    public PurchaseOrder(String address, Customer customer){
        this.address = address;
        this.customer = customer;
    }

    private PurchaseOrder(){
        this.address = null;
        this.customer = null;
    }
    /** Creates a new instance, will only be used by Jaxb. */
    private static PurchaseOrder newInstance() {
        return new PurchaseOrder();
    }

    public String getAddress() {
        return address;
    }

    public Customer getCustomer() {
        return customer;
    }
}

令人惊讶的是,这有效,并且在解组时会获得初始化的实例.您应该注意不要newInstance在代码的任何位置调用该方法,因为它将返回无效的实例.

Answer 3

您应该有一个JAXB的默认构造函数,以便能够实例化您的类.也许有一个我不知道的解决方法.

JAXB特别适合类似bean的类,允许通过调用它们上的setter来配置对象.