使用层次结构编组抽象类后删除 xsi:type

edw*_*ong 3 java xml jaxb

如果有人帮助我,那就太好了。所以我有一个结构:

<letters> <list> <name>Simon</name> <type>2</type> <passengerName>Johny</passengerName> <passengerSurname>Revelator</passengerSurname> </list> <list> <name>someName</name> <type>4</type> <fileURL>someUrl</fileURL> <specialNotes>specialNotes</specialNotes> </list> </letters>

对于这个结构,我编写了 Java 类,它们对它进行编组和解组:A 类包 com.edhex.testing;

import javax.xml.bind.annotation.*;
@XmlAccessorType(XmlAccessType.PROPERTY)
@XmlSeeAlso({B.class, C.class})
@XmlType
abstract public class A {
    int type;
    String name;

    @XmlElement
    public int getType() {
        return type;
    }

    public void setType(int type) {
        this.type = type;
    }

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

    public void setName(String name) {
        this.name = name;
    }
}
Run Code Online (Sandbox Code Playgroud)

班级信件:

@XmlRootElement
@XmlAccessorType(XmlAccessType.PROPERTY)
public class Letters {
    List<A> list;

    @XmlElement
    public List<A> getList() {
        return list;
    }

    public void setList(List<A> list) {
        this.list = list;
    }
}
Run Code Online (Sandbox Code Playgroud)

B类:

@XmlRootElement
@XmlType
@XmlAccessorType(XmlAccessType.PROPERTY)
public class B extends A {
    String fileURL;
    String specialNotes;
    @XmlElement
    public String getFileURL() {
        return fileURL;
    }

    public void setFileURL(String fileURL) {
        this.fileURL = fileURL;
    }
    @XmlElement
    public String getSpecialNotes() {
        return specialNotes;
    }

    public void setSpecialNotes(String specialNotes) {
        this.specialNotes = specialNotes;
    }
}
Run Code Online (Sandbox Code Playgroud)

C类:

@XmlRootElement
@XmlAccessorType(XmlAccessType.PROPERTY)

public class C extends A {
    String passengerName;
    String passengerSurname;
    @XmlElement
    public String getPassengerName() {
        return passengerName;
    }

    public void setPassengerName(String passengerName) {
        this.passengerName = passengerName;
    }
    @XmlElement
    public String getPassengerSurname() {
        return passengerSurname;
    }

    public void setPassengerSurname(String passengerSurname) {
        this.passengerSurname = passengerSurname;
    }
}
Run Code Online (Sandbox Code Playgroud)

一切看起来都很好,但在编组之后我确实喜欢这样:

public static void main(String[] args) {
        C c = new C();
        c.setName("Simon");
        c.setType(2);
        c.setPassengerName("Johny");
        c.setPassengerSurname("Revelator");
        B b = new B();
        b.setType(4);
        b.setFileURL("someUrl");
        b.setSpecialNotes("specialNotes");
        b.setName("someName");

        List<A> list = new ArrayList<A>(2);
        list.add(c);
        list.add(b);
        Letters letter = new Letters();
        letter.setList(list);
        try {
            File file = new File("C:\file.xml");
            JAXBContext jaxbContext = JAXBContext.newInstance(Letters.class);
            Marshaller jaxbMarshaller = jaxbContext.createMarshaller();
            jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
            jaxbMarshaller.marshal(letter, file);
            jaxbMarshaller.marshal(letter, System.out);
        } catch (JAXBException e) {
            e.printStackTrace();
        }
    }
Run Code Online (Sandbox Code Playgroud)

我得到这样的输出:

<letters>
    <list xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:type="c">
        <name>Lasha</name>
        <type>2</type>
        <passengerName>Johny</passengerName>
        <passengerSurname>Revelator</passengerSurname>
    </list>
    <list xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:type="b">
        <name>someName</name>
        <type>4</type>
        <fileURL>someUrl</fileURL>
        <specialNotes>specialNotes</specialNotes>
    </list>
</letters>
Run Code Online (Sandbox Code Playgroud)

问题:如何删除每个列表标记中的 xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:type="b" ?我知道 JAXB 会通知我它使用什么类型,因为最初映射使用另一种类型(A.class),但我不需要该信息。有人有什么想法吗?

bdo*_*han 5

@XmlElementRef如果您在属性上使用list,则该项目的元素将基于@XmlRootElement引用的类,然后您将无法获得属性xsi:type

@XmlRootElement
@XmlAccessorType(XmlAccessType.PROPERTY)
public class Letters {
    List<A> list;

    @XmlElementRef
    public List<A> getList() {
        return list;
    }

    public void setList(List<A> list) {
        this.list = list;
    }
}
Run Code Online (Sandbox Code Playgroud)