返回对象列表的Spring-Request方法显示错误

Question

@RestController
public class ApiController {
    @RequestMapping(value = "/", method = RequestMethod.GET)
    public String home() {
        return "<h1>Welcome to Spring REST API</h1>";
    }

    @RequestMapping(value = "/data", method = RequestMethod.GET, produces = MediaType.APPLICATION_JSON_UTF8_VALUE)
    public @ResponseBody List<UserData> data() {
        return new UserService().getData();
    }

}

public class UserService {

    private List<UserData> data;

    public UserService() {
        this.data = new ArrayList<UserData>();
        this.data.add(new UserData(1, "Leanne Graham", "bret", "[e-mail]", "Romaguera-Crona", "hildegard.org",
                "1-770-736-8031 x56442"));
        this.data.add(new UserData(2, "Ervin Howell", "Antonette", "[e-mail]", "Deckow-Crist", "anastasia.net",
                "010-692-6593 x09125"));
        this.data.add(new UserData(3, "Clementine Bauch", "Samantha", "[e-mail]", "Romaguera-Jacobson",
                "ramiro.info", "1-463-123-4447"));
        this.data.add(new UserData(4, "Patricia Lebsack", "Karianne", "[e-mail]", "Robel-Corkery",
                "kale.biz", "1-770-736-8037"));
        this.data.add(new UserData(5, "Chelsey Dietrich", "Kamren", "[e-mail]", "Keebler LLC",
                "demarco.info", "1-344-736-8031 x564"));
    }

    /**
     * @return the data
     */
    public List<UserData> getData() {
        return data;
    }

    /**
     * @param data the data to set
     */
    public void setData(List<UserData> data) {
        this.data = data;
    }

}

public class UserData {

    int id;
    String name, username, email, company, website, phone;

    public UserData(int id, String name, String username, String email, String company, String website, String phone) {
        this.id = id;
        this.name = name;
        this.username = username;
        this.email = email;
        this.company = company;
        this.website = website;
        this.phone = phone;
    }

}

会引发以下错误和异常：

路径为[]的上下文中servlet[dispatcherServlet]的servlet.Service（）引发异常[请求处理失败；嵌套异常为org.springframework.http.converter.httpmessageConversionException：类型定义错误:[简单类型，类com.example.springrest.userdata]；嵌套异常为com.fasterxml.jackson.databind.exc.invaliddefinitionException：未找到类com.example.springrest.userdata的序列化程序，也未发现创建BeanSerializer的属性（为避免异常，禁用

Answer 1

试试这个

控制器代码

@RestController
public class ApiController {

    @AutoWired
    UserService userservice;

    @RequestMapping(value = "/", method = RequestMethod.GET)
    public String home() {
        return "<h1>Welcome to Spring REST API</h1>";
    }

    @RequestMapping(value = "/data", method = RequestMethod.GET, produces = produces = MediaType.APPLICATION_JSON_VALUE)
    public List<UserData> data() {
        return new userservice.getData();
    }

}

服务代码

@Service
public class UserService {

    private List<UserData> data;

    public UserService() {
        this.data = new ArrayList<UserData>();
        this.data.add(new UserData(1, "Leanne Graham", "bret", "[e-mail]", "Romaguera-Crona", "hildegard.org",
                "1-770-736-8031 x56442"));
        this.data.add(new UserData(2, "Ervin Howell", "Antonette", "[e-mail]", "Deckow-Crist", "anastasia.net",
                "010-692-6593 x09125"));
        this.data.add(new UserData(3, "Clementine Bauch", "Samantha", "[e-mail]", "Romaguera-Jacobson",
                "ramiro.info", "1-463-123-4447"));
        this.data.add(new UserData(4, "Patricia Lebsack", "Karianne", "[e-mail]", "Robel-Corkery",
                "kale.biz", "1-770-736-8037"));
        this.data.add(new UserData(5, "Chelsey Dietrich", "Kamren", "[e-mail]", "Keebler LLC",
                "demarco.info", "1-344-736-8031 x564"));
    }

    /**
     * @return the data
     */
    public List<UserData> getData() {
        return data;
    }

    /**
     * @param data the data to set
     */
    public void setData(List<UserData> data) {
        this.data = data;
    }

}