jaxb - 如何使用错误的父/子模型解码 xml 消息

Question

我正在尝试将第 3 方 XML 有效负载解码到一个类中。问题是有效载荷具有父/子关系，并且根节点、父节点和子节点都具有相同的元素名称。这是有效负载的示例。

<?xml version="1.0" encoding="UTF-8"?>
<Directory>
    <id>2</id>
    <name>Media</name>
    <Directory>
        <id>5</id>
        <name>Default_Content</name>
        <Directory>
            <id>9</id>
            <name>Images</name>
        </Directory>
        <Directory>
            <id>8</id>
            <name>Icons</name>
        </Directory>
        <Directory>
            <id>6</id>
            <name>Additional_Content</name>
        </Directory>
    </Directory>
    <Directory>
        <id>12</id>
        <name>IC</name>
    </Directory>
</Directory>

所以我试图注释一个类，以便 JAXB/JAX-RS 可以将其解码为有用的东西。

我试过这样的东西

@XmlRootElement(name="Directory")
public class Directory {
    private int id;
    private String name;

    @XmlElement(name="Directory");
    private List<Directory> directories = new ArrayList<Directory>();
}

但是，可以预见的是，它会抛出 IllegalAnnotationException因为有两个同名的属性。

关于如何使用 JAXB/JAX-RS 干净地处理这种困惑的任何想法，还是我应该自己解析它？

Answer 1

简短回答

异常(exception)是由于字段/属性冲突。您可以注释属性(获取方法)或在您的类型上设置以下注释:

@XmlAccessorType(XmlAccessType.FIELD)
public class Directory {
  ...
}

长答案

JAXB 的默认访问类型是 PUBLIC_MEMBER这意味着 JAXB 将映射所有公共(public)字段(实例变量)和属性(get/set 方法)。

public class Foo {

    private String bar;

    public String getBar() {
        return bar;
    }

    public void setBar(String bar) {
        this.bar = bar;
    }

}

如果您注释一个字段:

public class Foo {

    @XmlAttribute
    private String bar;

    public String getBar() {
        return bar;
    }

    public void setBar(String bar) {
        this.bar = bar;
    }

}

那么 JAXB 会认为它有两个 bar属性映射并引发异常:

Exception in thread "main" com.sun.xml.bind.v2.runtime.IllegalAnnotationsException: 1 counts of IllegalAnnotationExceptions
Class has two properties of the same name "bar"
    this problem is related to the following location:
        at public java.lang.String example.Foo.getBar()
        at example.Foo
    this problem is related to the following location:
        at private java.lang.String example.Foo.bar
        at example.Foo

解决方法是对属性进行注解，并将 XmlAccessType 类型设置为 FIELD

@XmlAccessorType(XmlAccessType.FIELD)
public class Foo {

    @XmlAttribute
    private String bar;

    public String getBar() {
        return bar;
    }

    public void setBar(String bar) {
        this.bar = bar;
    }

}

您的型号

目录

import java.util.ArrayList;
import java.util.List;

import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement(name="Directory")
@XmlAccessorType(XmlAccessType.FIELD)
public class Directory {
    private int id;
    private String name;

    @XmlElement(name="Directory")
    private List<Directory> directories = new ArrayList<Directory>();

    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

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

    public List<Directory> getDirectories() {
        return directories;
    }

    public void setDirectories(List<Directory> directories) {
        this.directories = directories;
    }

}

演示

import java.io.File;
import javax.xml.bind.JAXBContext;
import javax.xml.bind.Marshaller;
import javax.xml.bind.Unmarshaller;

public class Demo {

    public static void main(String[] args) throws Exception {
        JAXBContext jc = JAXBContext.newInstance(Directory.class);

        Unmarshaller unmarshaller = jc.createUnmarshaller();
        Directory directory = (Directory) unmarshaller.unmarshal(new File("input.xml"));

        Marshaller marshaller = jc.createMarshaller();
        marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
        marshaller.marshal(directory, System.out);
    }

}