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我想将 Java 对象编组为 XML。对于这种方法，我将 JAXB Moxy 与外部 XML 绑定文件一起使用。

这是我想要编组为 XML 的示例类：

public class Customer {
  private String lastname;
  private String firstname;

  //getters and setters
}

使用我的外部绑定文件，我可以访问此类的值，以便获得以下 XML 输出：

<?xml version="1.0"?>
<customer>
  <firstname>Tony</firstname>
  <lastname>Stark</lastname>
</customer>

现在我想添加一个自定义标记，该标记具有未在 java 类中指定的自定义值。所以对于上面的 Customer 类，我想要一个这样的 XML 输出：

<?xml version="1.0"?>
<customer>
  <firstname>Tony</firstname>
  <lastname>Stark</lastname>
  <birthdate>01.01.1990</birthdate>
</customer>

Birthdate 不在 Customer 类中，我不想在其中添加它，因为该类是由脚本自动生成的。我的目标是在外部绑定文件中用我的自定义值定义生日。这甚至可以通过 JAXB MOXY 实现实现吗？希望可以有人帮帮我。

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1 回答 1

2

以下是您可以利用的一种方法XmlAdapter：

XmlAdapter（姓氏适配器）

import javax.xml.bind.annotation.XmlType;
import javax.xml.bind.annotation.adapters.XmlAdapter;

public class LastNameAdapter extends XmlAdapter<LastNameAdapter.AdaptedLastName, String> {

    @XmlType(propOrder={"lastname", "birthdate"})
    public static class AdaptedLastName {
        public String lastname;
        public String birthdate;
    }

    private String birthdate;

    public LastNameAdapter() {
    }

    public LastNameAdapter(String birthdate) {
        this.birthdate = birthdate;
    }

    @Override
    public String unmarshal(AdaptedLastName v) throws Exception {
        return v.lastname;
    }

    @Override
    public AdaptedLastName marshal(String v) throws Exception {
        AdaptedLastName adaptedLastName = new AdaptedLastName();
        adaptedLastName.lastname = v;
        adaptedLastName.birthdate = birthdate;
        return adaptedLastName;
    }

}

外部元数据 (oxm.xml)

<?xml version="1.0"?>
<xml-bindings
    xmlns="http://www.eclipse.org/eclipselink/xsds/persistence/oxm"
    package-name="forum19641824">
    <java-types>
        <java-type name="Customer">
            <xml-root-element/>
            <xml-type prop-order="firstname lastname"/>
            <java-attributes>
                <xml-element java-attribute="firstname"/>
                <xml-element java-attribute="lastname" xml-path=".">
                    <xml-java-type-adapter value="forum19641824.LastNameAdapter"/>
                </xml-element>
            </java-attributes>
        </java-type>
    </java-types>
</xml-bindings>

演示

import java.util.*;
import javax.xml.bind.*;
import org.eclipse.persistence.jaxb.JAXBContextProperties;

public class Demo {

    public static void main(String[] args) throws Exception {
        Map<String, Object> properties = new HashMap<String, Object>(1);
        properties.put(JAXBContextProperties.OXM_METADATA_SOURCE, "forum19641824/oxm.xml");
        JAXBContext jc = JAXBContext.newInstance(new Class[] {Customer.class}, properties);

        Customer customer = new Customer();
        customer.setFirstname("Tony");
        customer.setLastname("Stark");

        Marshaller marshaller = jc.createMarshaller();
        marshaller.setAdapter(new LastNameAdapter("01.01.1990"));
        marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
        marshaller.marshal(customer, System.out);
    }

}

输出

<?xml version="1.0" encoding="UTF-8"?>
<customer>
   <firstname>Tony</firstname>
   <lastname>Stark</lastname>
   <birthdate>01.01.1990</birthdate>
</customer>
于 2013-10-28T20:34:46.030 回答