Question

我有这个模型课程：

package org.myapp.model;
import java.util.Date;

import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;


@XmlRootElement(name="Message")
@XmlAccessorType(XmlAccessType.FIELD)
public class Message {

    public long id;
    public String message;
    public Date created;
    public String author;

    public Message() {

    }
    public Message(long id,String message, String author) {
        this.id = id;
        this.message = message;
        this.author = author;
        this.created = new Date();
    }
    @XmlElement
    public String getMessage() {
        return message;
    }
    public void setMessage(String message) {
        this.message = message;
    }
    @XmlElement
    public Date getCreated() {
        return created;
    }
    public void setCreated(Date created) {
        this.created = created;
    }
    @XmlElement
    public String getAuthor() {
        return author;
    }
    public void setAuthor(String author) {
        this.author = author;
    }


}

服务类别：

package org.myapp.services;


import java.util.ArrayList;
import java.util.Date;
import java.util.List;

import org.myapp.model.Message;



public class MessageService {


    public List<Message> getAllMessages(){
        Message msg1 = new Message(1L,"How are you?", "natalie");
        Message msg2 = new Message(2L,"How are you?", "amir");
        List<Message> msglist = new ArrayList<Message>();
        msglist.add(msg1);
        msglist.add(msg2);
        return msglist;
    }

}

资源类别：

package org.myapp.resource;

import java.util.List;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.PathParam;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;

import org.myapp.model.Message;
import org.myapp.services.MessageService;

@Path("messageresource")
public class MessageResource {

    MessageService messageService = new MessageService();

    @GET
    @Produces(MediaType.APPLICATION_XML)
    public List<Message> getMessage() {
        return   messageService.getAllMessages();  //"app chal rhi hai!";
    }

    @GET
    @Produces(MediaType.APPLICATION_XML)
    @Path("/{messageId}")
    public String getMessageID(@PathParam("messageId") String messageId) {
        Message message = new Message(1L,"How are you?", "natalie");
        return message.getMessage()+", "+messageId;
    }
}

我正在尝试以XML格式打印Service（MessageService）类中的数据。我认为是由于该类的方法（返回类型List<Message>）的返回值与Model（消息）类顶部的标记@XMLRootElement不一致而引起的。我尝试了其他 MediaType属性，但没有任何帮助。

当我访问此路径时：localhost:8080/messengerapp/webapi/messageresource
我收到此错误-Internal Server Error

我刚刚开始学习编写Web服务。我尝试了各种方法来解决此问题，但没有任何帮助。请帮助我理解并解决此问题。

Answer 1

尝试一下，

注释xml的根元素

@XmlRootElement (name="Messages")
public class MessageService implements Serializable{

    private List<Message> msglist = new ArrayList<Message>();

    public List<Message> getAllMessages(){
        Message msg1 = new Message(1L,"How are you?", "natalie");
        Message msg2 = new Message(2L,"How are you?", "amir");
        msglist.add(msg1);
        msglist.add(msg2);
        return msglist;
    }

    public void setAllMessages(List<Message> msglist){
        this.msglist = msglist;
    }

}

xml的结构

<Messages>  <!--  root element of xml  -->

    <Message>
        ...
    </Message>

    <Message>
        ...
    </Message>

</Messages>

获取内部服务器错误-@XmlRootElement

1 个答案: