希望对JAXB专家来说很简单:
我正在尝试编组一个不可变的类,它使不定义一个默认的无参数构造函数。我已经定义了一个XmlAdapter
实现,但它似乎没有被选中。我已经整理了一个简单的自包含示例,但仍然无法正常工作。谁能告诉我我做错了什么?
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; }
}
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。
答案 0 :(得分:7)
以下内容应该有所帮助:
FOO AS ROOT OBJECT
在类型级别指定@XmlJavaTypeAdapter
时,它仅适用于引用该类的字段/属性,而不适用于该类的实例是XML树中的根对象。这意味着您必须自己将Foo
转换为AdaptedFoo
,并在JAXBContext
而非AdaptedFoo
上创建Foo
。
<强>元帅强>
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();
}
}
}
<强> AdaptedFoo 强>
您需要在@XmlRootElement
课程中添加AdaptedFoo
注释。您可以从Foo
类中删除相同的注释。
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 NESTED OBJECT
当Foo
不是根对象时,一切都按照映射的方式工作。我已经扩展了你的模型来演示它是如何工作的。
<强>酒吧强>
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;
}
}
<强>演示强>
请注意,引导Foo
时,JAXB参考实现不允许您指定JAXBContext
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();
}
}
}
<强> input.xml中/输出强>
<?xml version="1.0" encoding="UTF-8"?>
<bar>
<foo name="Jane Doe" age="35"/>
</bar>
答案 1 :(得分:0)
我知道不是这种情况,但是如果在将@XmlJavaTypeAdapter用作字段时遇到此类错误,请检查是否指定了名称空间。您可能需要它。
对于我来说,这行不通:
@XmlElement(name = "Expiration")
@XmlJavaTypeAdapter(DateAdapter.class)
private Date expiration;
直到指定了名称空间:
@XmlElement(name = "Expiration", namespace="http://site/your.namespace")
@XmlJavaTypeAdapter(DateAdapter.class)
private Date expiration;