4

我有一些对象，比如说两个，A 和 B。这些对象来自同一个类。我需要使用 JAXB 编组这些对象，输出 XML 应采用以下形式：

<Root>
    <A>
        <ID> an id </ID>
    </A>
    <B>
        <ID> an id </ID>
    </B>
</Root>

<!-- Then all A and B attributes must be listed !-->
<A>
    <ID> an id </ID>
    <attribute1> value </attribute1>
    <attribute2> value </attribute2>
</A>
<B>
    <ID> an id </ID>
    <attribute1> value </attribute1>
    <attribute2> value </attribute2>
</B>

如何在 JAXB 中生成这种格式？任何帮助表示赞赏。

更新： 更具体地说，假设我们有这样的 Human 类：

@XmlRootElement
public class Human {
    private String name;
    private int age;
    private Integer nationalID;

    public String getName() {
        return name;
    }

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

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    public Integer getNationalID() {
        return nationalID;
    }

    public void setNationalID(Integer nationalID) {
        this.nationalID = nationalID;
    }
}

我们的主要课程是：

public class Main3 {

    public static void main(String[] args) throws JAXBException {
        Human human1 = new Human();
        human1.setName("John");
        human1.setAge(24);
        human1.setNationalID(Integer.valueOf(123456789));

        JAXBContext context = JAXBContext.newInstance(Human.class);
        Marshaller m = context.createMarshaller();
        m.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE);

        StringWriter stringWriter = new StringWriter();

        m.marshal(human1, stringWriter);

        System.out.println(stringWriter.toString());
    }

}

然后输出将是：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<human>
    <age>24</age>
    <name>John</name>
    <nationalID>123456789</nationalID>
</human>

现在我需要输出是这样的：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<human>
    <nationalID>123456789</nationalID>
</human>
<human>
    <nationalID>123456789</nationalID>
    <age>24</age>
    <name>John</name>
</human>

这将帮助我绘制一个没有属性（仅按 ID）的 XML 对象树，然后是树下的所有定义。这可能使用 JAXB 或任何其他实现吗？

4

2 回答 2

12

尝试这个：

import java.io.StringWriter;
import javax.xml.bind.Marshaller;

...

Object requestObject = ...  // This is the object that needs to be printed with indentation
Marshaller marshaller = ...
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE);
StringWriter stringWriter = new StringWriter();
marshaller.marshal(requestObject, stringWriter);

System.out.println(stringWriter.toString());
于 2012-05-10T12:46:51.593 回答
6 
package com.namasoft.dms.gui.common.utilities;

import java.io.StringWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.Random;

import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Marshaller;
import javax.xml.bind.annotation.XmlAccessorOrder;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlElementWrapper;
import javax.xml.bind.annotation.XmlID;
import javax.xml.bind.annotation.XmlIDREF;
import javax.xml.bind.annotation.XmlRootElement;

public class JAXB
{

    public static class Human
    {

        @XmlID
        String id;
        @XmlElement
        String name;
        @XmlElement
        int age;

        public Human()
        {
        }

        public Human(String name)
        {
            this.id = this.name = name;
            age = new Random().nextInt();
        }
    }

    @XmlRootElement
    public static class HumansList
    {
        @XmlElementWrapper(name = "humanObjects")
        @XmlElement(name = "human")
        List<Human> humanObjects = new ArrayList<>();

        @XmlIDREF
        @XmlElementWrapper(name = "humanIds")
        @XmlElement(name = "id")
        List<Human> humanIds = new ArrayList<>();

        void addHuman(Human human)
        {
            humanObjects.add(human);
            humanIds.add(human);
        }
    }

    public static void main(String[] args) throws JAXBException
    {
        HumansList list = new HumansList();
        Human parent1 = new Human("parent1");
        list.addHuman(parent1);
        Human child11 = new Human("child11");
        list.addHuman(child11);
        Human child12 = new Human("child12");
        list.addHuman(child12);

        Human parent2 = new Human("parent2");
        list.addHuman(parent2);
        Human child21 = new Human("child21");
        list.addHuman(child21);
        Human child22 = new Human("child22");
        list.addHuman(child22);

        JAXBContext context = JAXBContext.newInstance(HumansList.class);
        Marshaller m = context.createMarshaller();
        StringWriter xml = new StringWriter();
        m.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE);
        m.setProperty(Marshaller.JAXB_ENCODING, "UTF-8");

        m.marshal(list, xml);
        System.out.println(xml);
    }
}

输出将是

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<humansList>
    <humanObjects>
        <human>
            <id>parent1</id>
            <name>parent1</name>
            <age>-486071665</age>
        </human>
        <human>
            <id>child11</id>
            <name>child11</name>
            <age>920318383</age>
        </human>
        <human>
            <id>child12</id>
            <name>child12</name>
            <age>-1355111983</age>
        </human>
        <human>
            <id>parent2</id>
            <name>parent2</name>
            <age>-314154051</age>
        </human>
        <human>
            <id>child21</id>
            <name>child21</name>
            <age>983544381</age>
        </human>
        <human>
            <id>child22</id>
            <name>child22</name>
            <age>748252616</age>
        </human>
    </humanObjects>
    <humanIds>
        <id>parent1</id>
        <id>child11</id>
        <id>child12</id>
        <id>parent2</id>
        <id>child21</id>
        <id>child22</id>
    </humanIds>
</humansList>
于 2012-05-13T08:53:57.370 回答