JAXB 和构造函数

constructor java jaxb creation

2022-09-01 12:41:47

我开始学习JAXB，所以我的问题可能非常愚蠢。现在我有类，想要生成XML架构。按照这个指令，我得到异常

非法注释例外情况 ...没有无参数默认构造函数。

是的。我的类没有默认的无参数构造函数。这太容易了。我有带有包可见构造函数/最终方法的类，并且带有参数。我该怎么办 - 创建一些特定的mometo/构建器类或将我的构造函数指定给JAXB（以什么方式？）？谢谢。

答案 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 arg 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;
    }

}

以及要在它们之间进行转换的 XML 适配器：

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注释：

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

答案 2

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

@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，这是有效的，并且在取消编组时您将获得初始化的实例。您应该注意不要在代码上的任何位置调用该方法，因为它将返回无效的实例。newInstance