jersey-2.0 - Jersey 在尝试返回 XML 响应时返回 500

Question

我正在尝试使用基于 this 的 Jersey 2.12 创建我自己的 RESTful WS 应用程序文章。我想根据从 url 传递的 id 返回一个类的 XML 表示，但是，从高级 Rest 客户端应用程序(谷歌浏览器应用程序)或浏览器尝试时，我收到了 500 响应代码。以下是详细信息:

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
    xmlns="http://java.sun.com/xml/ns/javaee" 
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-
    app_3_0.xsd" id="WebApp_ID" version="3.0">
  <display-name>WS_RESTful_Practice</display-name>
  <servlet>
    <servlet-name>Jersey REST Service</servlet-name>
    <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
    <!-- Register resources and providers under com.vogella.jersey.first package. -->
    <init-param>
        <param-name>jersey.config.server.provider.packages</param-name>
        <param-value>test.services</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
 </servlet>
 <servlet-mapping>
    <servlet-name>Jersey REST Service</servlet-name>
    <url-pattern>/rest/*</url-pattern>
 </servlet-mapping>
</web-app> 

TestRestModel.java

package test.model;

import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement
public class TestRestModel{

    /**
     * 
     */
    private static final long serialVersionUID = -8391589100962515747L;
    private String name;
    private String content;

    public TestRestModel(String name, String content){
        this.name = name;
        this.content = content;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getContent() {
        return content;
    }

    public void setContent(String content) {
        this.content = content;
    }
}

TestResource.java

package test.services;

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 javax.ws.rs.core.Response;

import test.dao.TestModelDao;
import test.model.TestRestModel;

@Path("/test")
public class TestResource {

    @GET
    @Path("{id}")
    public Response getModel(@PathParam("id") String id){
        return Response.ok().entity(TestModelDao.instance.getModel().get(id)).build();
    }
}

TestModelDao.java

package test.dao;

import java.util.HashMap;
import java.util.Map;
import test.model.TestRestModel;

public enum TestModelDao {
    instance;

    private Map<String, TestRestModel> container = new HashMap<String, TestRestModel>();

    private TestModelDao(){

        TestRestModel model = new TestRestModel("a", "this is first");
        container.put("1", model);
        model = new TestRestModel("b", "this is second");
        container.put("2", model);
        model = new TestRestModel("c", "this is third");
        container.put("3", model);
    }

    public Map<String, TestRestModel> getModel(){
        return container;
    }
}

我对 Jersey 和 REST 完全陌生。而且我还不知道如何从 Jersey 记录错误。

Answer 1

当您没有为 JAXB bean 提供默认的无参数构造函数时，就会发生这种情况。因此，在您的情况下，您应该通过添加一个来修改类:

  public TestRestModel(){

    }

这是由于 JSR-222 中的要求所致:

existing types, authored by users, are required to provide a no arg constructor. The no arg constructor is used by an unmarshaller during unmarshalling to create an instance of the type.