JAXB編組沒有嵌套的對象與新澤西

Question

import com.sun.jersey.api.client.Client; 
import com.sun.jersey.api.client.ClientResponse; 
import com.sun.jersey.api.client.WebResource; 

public class App { 
    public static void main(String[] args){ 
     Client client = Client.create(); 
     WebResource webResource = client.resource("https://mywebsite.com/getDevices"); 
     ClientResponse response = webResource.accept("application/xml").get(
       ClientResponse.class); 
     System.out.println(response.getEntity(Devices.class)); 
    } 
} 

Devices.java

import java.util.ArrayList; 
import java.util.List; 
import javax.xml.bind.annotation.XmlAccessType; 
import javax.xml.bind.annotation.XmlAccessorType; 
import javax.xml.bind.annotation.XmlElement; 
import javax.xml.bind.annotation.XmlRootElement; 
import javax.xml.bind.annotation.XmlType; 

@XmlAccessorType(XmlAccessType.FIELD) 
@XmlType(namespace="myNamespace", propOrder = {"deviceList"}) 
@XmlRootElement(name = "Entries", namespace="myNamespace") 
public class Devices { 

    @XmlElement(name = "Entry", namespace="myNamespace") 
    protected List<Devices.Device> deviceList; 

    public List<Devices.Device> getEntry() { 
     if (deviceList == null) { 
      deviceList = new ArrayList<Devices.Device>(); 
     } 
     return this.deviceList; 
    } 

    @Override 
    public String toString() { 
     return deviceList.toString(); 
    } 

    @XmlAccessorType(XmlAccessType.FIELD) 
    @XmlType(propOrder = {"devicename"}) 
    public static class Device { 
     String devicename; 

     public String getDevicename() { 
      return devicename; 
     } 

     public void setDevicename(String value) { 
      this.devicename = value; 
     } 

     @Override 
     public String toString() { 
      return devicename; 
     } 
    } 
} 

樣品從XML Web服務

<Entries xmlns="myNamespace"> 
    <Entry><devicename xmlns="myNamespace">Device1</devicename></Entry> 
    <Entry><devicename xmlns="myNamespace">Device2</devicename></Entry> 
</Entries> 

回到它似乎在數據正確地拉，但每個返回null設備名稱。

Answer 1

而不是你能怎麼你正在映射您的命名空間限定使用包級別@XmlSchema標註上一個名爲package-info類，如下所示：

package-info.java

@XmlSchema(
    namespace = "myNamespace", 
    elementFormDefault = XmlNsForm.QUALIFIED) 
package your.pkg; 

import javax.xml.bind.annotation.XmlNsForm; 
import javax.xml.bind.annotation.XmlSchema; 

更多信息

• http://blog.bdoughan.com/2010/08/jaxb-namespaces.html