XmlJavaTypeAdapter沒有檢測

Question

希望一個容易的JAXB專家：

我試圖獲取一個不可變的類，它不定義一個默認的無參數的構造函數。我已經定義了一個XmlAdapter實現，但它似乎沒有被拾取。我已經制定了一個簡單的，自成一體的例子，但仍然失敗。任何人都可以提醒我做錯了什麼？

不可變類

@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; } 
} 

適配器和值類型

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; } 
} 

的Marshaller

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(); 
    } 
    } 
} 

堆棧跟蹤

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) 

請注意，我用的JDK 1.7.0_05。

Answer 1

以下應該有所幫助：

FOO AS根對象

當@XmlJavaTypeAdapter在它僅適用於域/屬性引用該類類型級別指定，而不是在這個類的一個實例是您的XML樹中的根對象。這意味着您必須自己將Foo轉換爲AdaptedFoo，並在AdaptedFoo上創建JAXBContext，而不是Foo。

元帥

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

您將需要一個@XmlRootElement註釋添加到AdaptedFoo類。您可以從Foo類中刪除相同的註釋。

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; 
    } 
} 

---

FOO AS嵌套對象

當Foo不是根對象一切正常，你把它映射的方式。我已經擴展了您的模型以演示這將如何工作。

酒吧

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

注意，JAXB參考實現不會讓你自舉JAXBContext時指定Foo類。

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(); 
     } 
    } 
} 

的input.xml /輸出

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