Question

为了使用JAXB编组一个长基本类型，我使用了@ XmlJavaTypeAdapter注释，它将非String类型调整为String。即使它为长类型抛出错误。为什么会这样？如何对long id属性进行编组？

User.java

class User {
    @XmlID
    @XmlJavaTypeAdapter(WSLongAdapter.class)
    private long id;
    // Other variables
    // Getter & Setter method
}

WSLongAdapter.java

    public class WSLongAdapter extends XmlAdapter<String, Long>{
        @Override
        public String marshal(Long id) throws Exception {
            if(id==null) return "" ;
            return id.toString();
        }
        @Override
        public Long  unmarshal(String id) throws Exception {
        return  Long.parseLong(id);
        }
     }

MarshallTest.java

public static void main(String[] args) {
    try{
        JAXBContext jaxbContext= JAXBContext.newInstance(User.class);
        Marshaller marshaller = jaxbContext.createMarshaller();
        marshaller.setProperty(Marshaller.JAXB_FRAGMENT, true);
        OutputStreamWriter writer = new OutputStreamWriter(System.out);
        // Manually open the root element
        writer.write("<user>");
        // Marshal the objects out individually
        marshaller.marshal(new User(), writer);
        // Manually close the root element
        writer.write("</user>");
        writer.close();
    }
    catch (Exception e) {
        e.printStackTrace();
    }
}

错误：

  com.sun.xml.bind.v2.runtime.IllegalAnnotationsException: 4 counts of IllegalAnnotationExceptions
Adapter com.v4common.shared.util.other.WSLongAdapter is not applicable to the field type long. 
    this problem is related to the following location:
        at @javax.xml.bind.annotation.adapters.XmlJavaTypeAdapter(type=class javax.xml.bind.annotation.adapters.XmlJavaTypeAdapter$DEFAULT, value=class com.v4common.shared.util.other.WSLongAdapter)
        at private long com.v4common.shared.beans.usermanagement.User.id
        at com.v4common.shared.beans.usermanagement.User
Property "id" has an XmlID annotation but its type is not String.
    this problem is related to the following location:
        at private long com.v4common.shared.beans.usermanagement.User.id
        at com.v4common.shared.beans.usermanagement.User
There are two properties named "id"

Answer 1

在注释中指定基本类：

@XmlJavaTypeAdapter(type=long.class, value=WSLongAdapter.class)

Answer 2

以下可能有效：

class User {
    @XmlID
    @XmlJavaTypeAdapter(WSLongAdapter.class)
    @XmlElement(type=Long.class)
    private long id;
    // Other variables
    // Getter & Setter method
}

Answer 3

创建使用@XmlID注释的新getter返回String

@XmlAccessorType(XmlAccessType.FIELD)
class User {

    private long id;

    @XmlID
    public String getReferenceId() {
        return Long.toString(id);
    }

}

此解决方案的唯一缺点是序列化XML中存在其他标记（在本例中为<referenceId>）。我尝试通过使用@XmlTransient注释getter来删除它，但后来我收到错误通知引用类中不存在ID。

Answer 4

只需替换

 @XmlElement(type=Long.class)
private long id;

与

    @XmlSchemaType(name = "long")
protected Long id;

使用JAXB对长基本类型进行编组

4 个答案: