如何使用java注释读取xsi:type

Question

我想将基于 jaxb 的 xml 文件读入我的面向对象结构。可以说这是我的 xml 文件：    <?xml version="1.0" encoding="utf-8" standalone="yes"?>    <children xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">        <child xsi:type="girl">            <age>12</age>            <isdancing>true</isdancing>        </child>        <child xsi:type="boy">            <age>10</age>            <issoccerplayer>true</issoccerplayer>        </child>    </children>Children是某种包含多个子元素的包装元素。孩子可以是 xsi : type 指定的男孩或女孩。这两个类有一些共同的元素（如age）和一些不同（排除）的元素（如isdancing或issoccerplayer）要读取文件，我有这个方法：    public static void main( String[] args ) throws JAXBException    {        JAXBContext jaxbContext;        jaxbContext = JAXBContext.newInstance(Children.class);                     Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();        File file = new File("C:/test.xml");        if (!file.exists()) System.out.println("File does not exist");        Children children = (Children) jaxbUnmarshaller.unmarshal(file);        System.out.println(children.toString());    }我的孩子们的班级是这样的：    @XmlRootElement(name="children")    @XmlAccessorType(XmlAccessType.FIELD)    public class Children {        @XmlElement(name="child")        private List<Child> childrenList;        public List<Child> getChildren() { return childrenList; }        public void setChildren(List<Child> children) {this.childrenList = children;}    @Override        public String toString() {            return ReflectionToStringBuilder.toString(this);        }    }我的孩子班看起来像这样：    @XmlAccessorType(XmlAccessType.FIELD)    public class Child {    @XmlAttribute(name="xsi:type")    private XsiType xsiType;    private int age;    @XmlElement(name = "isdancing")    private boolean isDancing;        }    }我现在的问题是，输出正常，但 Child-class 的元素 xsiType 始终为 null，否则最终会出现 IllegalAnnotationExceptions，这与 XmlTest.model.Child.xsiType 相关所以我预计设置任何类型的 @Xml-Annotation 都会出现错误。有人可以帮我找出错误吗？目标是迭代孩子列表并在运行时（基于 xsiType）决定这是女孩还是男孩。

Answer 1

你不需要你的XsiType课。您可以直接使用String。

在你的Child类中xsiType，属性应该如下所示。

@XmlAttribute(name = "type", namespace = " 
private String xsiType;

注意：在@XmlAttribute注释中

• 使用name = "type"（不带前缀xsi:）

• 指定namespaceXML 中给定的参数xmlns:xsi="..."

---

顺便说一句： 您最好使用常量，
而不是键入字符串。所以你的改进代码会是这样的："http://www.w3.org/2001/XMLSchema-instance"XMLConstants.W3C_XML_SCHEMA_INSTANCE_NS_URI

@XmlAttribute(name = "type", namespace = XMLConstants.W3C_XML_SCHEMA_INSTANCE_NS_URI)
private String xsiType;

Answer 2

xsi 类型通常用于表达对具体类型的引用。Jaxb 可以使用 xsi 类型，无需进一步的解决方法。

创建一个Boy和 一个Girl扩展的类Children。（您可能需要使用 调整类型名称@XmlType）。这样，所有具有的元素都xsi:type=Girl将绑定到该类Girl

@XmlAccessorType(XmlAccessType.FIELD)

@XmlSeeAlso({ Boy.class, Girl.class }) // Either use @XmlSeeAlso to register classes in the JaxbContext

                                       //  or add them to the context directly

public class Child {

    private int age;

    @XmlElement(name = "isdancing")

    private boolean isDancing;

    @XmlElement(name = "issoccerplayer")

    private boolean isSoccerPlayer;

    // Getter and setter for all fields

}

@XmlType(name = "boy") // can be omitted if default value matches with the default value

public class Boy extends Child {

}

@XmlType(name = "girl")

public class Girl extends Child {

}

完整的独立示例：

package jaxb;

import java.io.File;

import java.io.StringReader;

import java.util.List;

import javax.xml.bind.JAXBContext;

import javax.xml.bind.JAXBException;

import javax.xml.bind.Unmarshaller;

import javax.xml.bind.annotation.XmlAccessType;

import javax.xml.bind.annotation.XmlAccessorType;

import javax.xml.bind.annotation.XmlElement;

import javax.xml.bind.annotation.XmlRootElement;

import javax.xml.bind.annotation.XmlSeeAlso;

import javax.xml.bind.annotation.XmlType;

public class Inheritance {

    public static void main(String[] args) throws JAXBException {

        JAXBContext jaxbContext;

        jaxbContext = JAXBContext.newInstance(Children.class);

        Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();

        String x = "<?xml version=\"1.0\" encoding=\"utf-8\" standalone=\"yes\"?>\r\n"

                + "    <children xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\">\r\n"

                + "        <child xsi:type=\"girl\">\r\n" + "            <age>12</age>\r\n"

                + "            <isdancing>true</isdancing>\r\n" + "        </child>\r\n"

                + "        <child xsi:type=\"boy\">\r\n" + "            <age>10</age>\r\n"

                + "            <issoccerplayer>true</issoccerplayer>\r\n" + "        </child>\r\n" + "    </children>";

        Children children = (Children) jaxbUnmarshaller.unmarshal(new StringReader(x));

        System.out.println(children.getChildren().toString());

    }

    @XmlRootElement(name = "children")

    @XmlAccessorType(XmlAccessType.FIELD)

    public static class Children {

        @XmlElement(name = "child")

        private List<Child> childrenList;

        public List<Child> getChildren() {

            return childrenList;

        }

        public void setChildren(List<Child> children) {

            this.childrenList = children;

        }

    }

    @XmlAccessorType(XmlAccessType.FIELD)

    @XmlSeeAlso({ Boy.class, Girl.class })

    public static class Child {

        private int age;

        @XmlElement(name = "isdancing")

        private boolean isDancing;

        @XmlElement(name = "issoccerplayer")

        private boolean isSoccerPlayer;

        // Getter and setter for all fields

    }

    @XmlType(name = "boy")

    public static class Boy extends Child {

    }

    @XmlType(name = "girl")

    public static class Girl extends Child {

    }

}

Answer 3

第二种方法的干净解决方案（基于单独的类文件）：

public class App

{

    public static void main(String[] args) throws JAXBException

    {

        JAXBContext jaxbContext = JAXBContext.newInstance(Children.class);             

        Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();

        File file = new File("C:/test2.xml");

        Children children = (Children) jaxbUnmarshaller.unmarshal(file);

        for (Child c : children.getChildren()) {

            if (c instanceof Boy) {

                System.out.println(((Boy)c).toString());

            } else if (c instanceof Girl){

                System.out.println(((Girl)c).toString());

            }

        }

    }

}

Children.java

@XmlRootElement(name="children")

@XmlAccessorType(XmlAccessType.FIELD)

public class Children {

    @XmlElement(name="child")

    private List<Child> childrenList;

    public List<Child> getChildren() { return childrenList; }

    public void setChildren(List<Child> children) {this.childrenList = children;}

    @Override

    public String toString() { return ReflectionToStringBuilder.toString(this); }

}

Boy.java

@XmlType(name="boy")

public class Boy extends Child {

    @XmlElement(name = "issoccerplayer")

    private boolean isSoccerPlayer;

    public boolean isSoccerPlayer() { return isSoccerPlayer; }

    public void setSoccerPlayer(boolean isSoccerPlayer) { this.isSoccerPlayer = isSoccerPlayer; }

    @Override

    public String toString() { return ReflectionToStringBuilder.toString(this); }

}

Girl.java

@XmlType(name="girl")

public class Girl extends Child {

    @XmlElement(name = "isdancing")

    private boolean isDancing;

    public boolean isDancing() { return isDancing; }

    public void setDancing(boolean isDancing) { this.isDancing = isDancing; }

    @Override

    public String toString() { return ReflectionToStringBuilder.toString(this); }

}

Child.java

@XmlAccessorType(XmlAccessType.FIELD)

@XmlSeeAlso({ Boy.class, Girl.class }) 

public abstract class Child {

    private int age;

    public int getAge() { return age; }

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

}

输出应该是：

de.home.myproject.XmlTest.model.Girl@12edcd21[isDancing=true,age=12]

de.home.myproject.XmlTest.model.Boy@27bc2616[isSoccerPlayer=true,age=10]