java - 未检测到 XmlJavaTypeAdapter

Question

希望对 JAXB 专家来说是一个简单的方法:

我正在尝试编码(marshal)一个不定义默认无参数构造函数的不可变类。我已经定义了一个 XmlAdapter 实现，但它似乎没有被拾取。我整理了一个简单的独立示例，但仍然无法正常工作。谁能告诉我做错了什么？

不可变类

@XmlJavaTypeAdapter(FooAdapter.class)
@XmlRootElement
public class Foo {
  private final String name;
  private final int age;

  public Foo(String name, int age) {
    this.name = name;
    this.age = age;
  }

  public String getName() { return name; }
  public int getAge() { return age; }
}

适配器和值类型

public class FooAdapter extends XmlAdapter<AdaptedFoo, Foo> {
  public Foo unmarshal(AdaptedFoo af) throws Exception {
    return new Foo(af.getName(), af.getAge());
  }

  public AdaptedFoo marshal(Foo foo) throws Exception {
    return new AdaptedFoo(foo);
  }
}

class AdaptedFoo {
  private String name;
  private int age;

  public AdaptedFoo() {}

  public AdaptedFoo(Foo foo) {
    this.name = foo.getName();
    this.age = foo.getAge();
  }

  @XmlAttribute
  public String getName() { return name; }
  public void setName(String name) { this.name = name; }

  @XmlAttribute
  public int getAge() { return age; }
  public void setAge(int age) { this.age = age; }
}

编码器

public class Marshal {
  public static void main(String[] args) {
    Foo foo = new Foo("Adam", 34);

    try {
      JAXBContext jaxbContext = JAXBContext.newInstance(Foo.class);
      Marshaller jaxbMarshaller = jaxbContext.createMarshaller();

      // output pretty printed
      jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

      jaxbMarshaller.marshal(foo, System.out);              
    } catch (JAXBException e) {
      e.printStackTrace();
    }   
  }
}

堆栈跟踪

com.sun.xml.internal.bind.v2.runtime.IllegalAnnotationsException: 1 counts of IllegalAnnotationExceptions
Foo does not have a no-arg default constructor.
        this problem is related to the following location:
                at Foo

        at com.sun.xml.internal.bind.v2.runtime.IllegalAnnotationsException$Builder.check(IllegalAnnotationsException.java:91)
        at com.sun.xml.internal.bind.v2.runtime.JAXBContextImpl.getTypeInfoSet(JAXBContextImpl.java:451)
        at com.sun.xml.internal.bind.v2.runtime.JAXBContextImpl.<init>(JAXBContextImpl.java:283)
        at com.sun.xml.internal.bind.v2.runtime.JAXBContextImpl.<init>(JAXBContextImpl.java:126)
        at com.sun.xml.internal.bind.v2.runtime.JAXBContextImpl$JAXBContextBuilder.build(JAXBContextImpl.java:1142)
        at com.sun.xml.internal.bind.v2.ContextFactory.createContext(ContextFactory.java:130)
        at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
        at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
        at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
        at java.lang.reflect.Method.invoke(Method.java:601)
        at javax.xml.bind.ContextFinder.newInstance(ContextFinder.java:248)
        at javax.xml.bind.ContextFinder.newInstance(ContextFinder.java:235)
        at javax.xml.bind.ContextFinder.find(ContextFinder.java:445)
        at javax.xml.bind.JAXBContext.newInstance(JAXBContext.java:637)
        at javax.xml.bind.JAXBContext.newInstance(JAXBContext.java:584)
        at Marshal2.main(Marshal2.java:11)

请注意，我使用的是 JDK 1.7.0_05。

Answer 1

以下应该有所帮助:

作为根对象的 FOO

当 @XmlJavaTypeAdapter在类型级别指定它仅适用于引用该类的字段/属性，而不是当该类的实例是 XML 树中的根对象时。这意味着您必须转换 Foo至 AdaptedFoo自己，并创建 JAXBContext在 AdaptedFoo而不是 Foo .

编码(marshal)

package forum11966714;

import javax.xml.bind.*;

public class Marshal {
    public static void main(String[] args) {
      Foo foo = new Foo("Adam", 34);

      try {
        JAXBContext jaxbContext = JAXBContext.newInstance(AdaptedFoo.class);
        Marshaller jaxbMarshaller = jaxbContext.createMarshaller();

        // output pretty printed
        jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

        jaxbMarshaller.marshal(new AdaptedFoo(foo), System.out);              
      } catch (JAXBException e) {
        e.printStackTrace();
      }   
    }
  }

AdaptedFoo

您需要添加 @XmlRootElement AdaptedFoo 的注释类(class)。您可以从 Foo 中删除相同的注释类。

package forum11966714;

import javax.xml.bind.annotation.*;

@XmlRootElement
class AdaptedFoo {
    private String name;
    private int age;

    public AdaptedFoo() {
    }

    public AdaptedFoo(Foo foo) {
        this.name = foo.getName();
        this.age = foo.getAge();
    }

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

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

    @XmlAttribute
    public int getAge() {
        return age;
    }

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

---

作为嵌套对象的 FOO

当 Foo不是根对象，一切都按照您映射的方式工作。我已经扩展了您的模型来演示它是如何工作的。

栏

package forum11966714;

import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement
public class Bar {

    private Foo foo;

    public Foo getFoo() {
        return foo;
    }

    public void setFoo(Foo foo) {
        this.foo = foo;
    }

}

演示

请注意，JAXB 引用实现不会让您指定 Foo引导 JAXBContext 时的类.

package forum11966714;

import java.io.File;

import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Marshaller;
import javax.xml.bind.Unmarshaller;

public class Demo {
    public static void main(String[] args) {
        try {
            JAXBContext jaxbContext = JAXBContext.newInstance(Bar.class);

            Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
            File xml = new File("src/forum11966714/input.xml");
            Bar bar = (Bar) jaxbUnmarshaller.unmarshal(xml);

            Marshaller jaxbMarshaller = jaxbContext.createMarshaller();
            jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
            jaxbMarshaller.marshal(bar, System.out);
        } catch (JAXBException e) {
            e.printStackTrace();
        }
    }
}

input.xml/输出

<?xml version="1.0" encoding="UTF-8"?>
<bar>
    <foo name="Jane Doe" age="35"/>
</bar>