在spring data rest中使用自定义控制器将URI转换为实体？答案

【问题标题】：converting URI to entity with custom controller in spring data rest?在spring data rest中使用自定义控制器将URI转换为实体？
【发布时间】：2017-09-21 22:23:08
【问题描述】：

我有一个这样的 jpa 实体。

@Entity
@Table(name = "location")
@Data
public class Location {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private long id;

    @Column(name = "LOCATION_ID", unique = true)
    @NotEmpty(message = "Please Enter Location ID")

    private String name;

    @Column(name = "LOCATION_DESCRIPTION")
    @NotEmpty(message = "Please Enter Location Description")
    private String description;

    @ManyToOne
    @NotNull(message = "Please Choose a Building")


    Building building;



    @Version
    Long version;





}

和这样的存储库。

public interface LocationRepository extends PagingAndSortingRepository<Location, Long> {  
    Location findByName(@Param("name") String name);

}

我正在使用 spring data rest 我可以通过提供以下有效负载来使用 rest api 创建位置

{

"name":"adminxxxxx","description":"adminxxx" , "building": "http://localhost:8080/buildings/2"
}

现在我正在尝试编写将保留实体的自定义控制器。这是我的自定义控制器

@ExposesResourceFor(Location.class)
@RepositoryRestController
@BasePathAwareController
public class LocationController {

    @Autowired
    LocationRepository locationDao;

    @Autowired
    LocationResourceAssembler resourceAssembler;

    @Value("${buildings.error.messages.uniqueconstraintviolation}")
    String uniqueConstrainMessage;
    static final String TAG = LocationController.class.getSimpleName();

    @RequestMapping(value="locations",method = org.springframework.web.bind.annotation.RequestMethod.POST)
    public ResponseEntity<?> save(@RequestBody @Valid Location location) {

        try {
            location = locationDao.save(location);
            LocationResource b = resourceAssembler.toResource(location);
            return ResponseEntity.ok().body(b);
        } catch (DataIntegrityViolationException e) {

            if (locationAlreadyExists(location.getName()))
                throw new LocationAlreadyExistException(uniqueConstrainMessage, location);

            else
                throw new RuntimeException("Some Error Occured");
        }

    }

我收到了这个错误

exception is com.fasterxml.jackson.databind.JsonMappingException: Can not construct instance of com.alamdar.model.Building: no String-argument constructor/factory method to deserialize from String value (&#39;http://localhost:8080/buildings/2&#39;)
 at [Source: java.io.PushbackInputStream@5d468b16; line: 3, column: 60] (through reference chain: com.alamdar.model.Location[&quot;building&quot;])</div></body></html>

有人可以帮忙吗？

【问题讨论】：

这个stackoverflow.com/q/41324078/5380322 可以提供帮助；）
你为什么要编写一个自定义控制器来进行保存？
我写自定义控制器的原因是因为我需要在持久化实体之前编写一些自定义逻辑。
Resolving entity URI in custom controller (Spring HATEOAS)的可能重复

标签： spring spring-data-rest

【解决方案1】：

我不确定您为什么要编写自定义控制器，但问题似乎是您没有默认的无参数构造函数，因此 Jackson 无法实例化实例。

这是因为你使用的是 Lombok 的 @Data 注解：

https://projectlombok.org/features/Data.html

您还应该使用 @NoArgsConstructor 注释您的类以生成默认的无参数构造函数：

@Entity
@Table(name = "location")
@Data
@NoArgsConstructor
public class Location {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private long id;

    @Column(name = "LOCATION_ID", unique = true)
    @NotEmpty(message = "Please Enter Location ID")

    private String name;

    @Column(name = "LOCATION_DESCRIPTION")
    @NotEmpty(message = "Please Enter Location Description")
    private String description;

    @ManyToOne
    @NotNull(message = "Please Choose a Building")
    Building building;

    @Version
    Long version;
}

【讨论】：