FindByUUID（）使用Spring Data的JPA存储库 [英] FindByUUID() using Spring Data's JPA Repository

问题描述

由于某种原因，我无法为此找到合适的答案。我有以下简单的实体：

$ p $ @Entity
@Table（name =simple_entity）
@Access（AccessType.FIELD）
public class SimpleEntity {

@Id
@GeneratedValue（strategy = GenerationType.AUTO）
protected Long id;

@Column（unique = true，updatable = false）
保护UUID uuid;

@PrePersist
protected void onCreateAbstractBaseEntity（）{
this.uuid = UUID.randomUUID（）;
}

public Long getId（）{
return this.id;
}

public UUID getUuid（）{
return this.uuid; Spring数据JPA和Hibernate在我的MySQL中创建了一切正确的数据数据库。但是，当我尝试使用我的JPARepository实现来使用它的uuid搜索一个项目时，即使它在数据库上执行find查询（我可以在调试器中看到它），它也找不到任何东西。这是我的JPARepository实现：
$ b $ pre $ public interface SimpleEntityRepository扩展JpaRepository< SimpleEntity，Long> {
SimpleEntity findOneByUuid（UUID uuid）;

$ / code>

这是调用此方法的控制器。

  @Controller 
 @RequestMapping（/ simple_entity）
公共类SimpleEntityController {
 
 @Autowired 
私人SimpleEntityRepository存储库; 
 
 @RequestMapping（method = RequestMethod.GET，value =/ {simpleEntityId}，产生= MediaType.APPLICATION_JSON_VALUE）
 public ResponseEntity< FileDatastore> getSimpleEntity（@PathVariable UUID simpleEntityId）{
 SimpleEntity record = this.repository.findOneByUuid（simpleEntityId）; 
 HttpHeaders headers = new HttpHeaders（）; 
 
 HttpStatus status =（record！= null）？ HttpStatus.OK：HttpStatus.NOT_FOUND; 
 
返回新的ResponseEntity<>（记录，标题，状态）; 
} 
  

我缺少什么？

感谢您的帮助！

解决方案

尝试注释您的 UUID @ org.hibernate.annotations.Type（type =org.hibernate.type.UUIDCharType）

或$ @ org.hibernate.annotations.Type（type =org.hibernate.type.UUIDBinaryType）

I由于MSB / LSB以二进制格式与 UUID 交换，所以在查询带有 UUID 的数据库时遇到类似的问题。我们解决了将数据作为字符串处理的问题，并在转换之前进行了必要的转换。

for some reason I have not being able to find a suitable answer for this. I have the following simple entity:

@Entity
@Table(name = "simple_entity")
@Access(AccessType.FIELD)
public class SimpleEntity {

  @Id
  @GeneratedValue(strategy = GenerationType.AUTO)
  protected Long id;

  @Column(unique = true, updatable = false)
  protected UUID uuid;

  @PrePersist
  protected void onCreateAbstractBaseEntity() {
      this.uuid = UUID.randomUUID();
  }

  public Long getId() {
      return this.id;
  }

  public UUID getUuid() {
      return this.uuid;
  }
}

Spring Data JPA with Hibernate creates everything correctly in my MySQL database. However, when I try to use my JPARepository implementation to search for an item using its uuid, it never finds anything, even though it executes the find query on the DB (which I can see in my debugger). Here is my JPARepository implementation:

public interface SimpleEntityRepository extends JpaRepository<SimpleEntity, Long> {
      SimpleEntity findOneByUuid(UUID uuid);
}

Here is the controller that calls this method.

@Controller
@RequestMapping("/simple_entity")
public class SimpleEntityController {

@Autowired
private SimpleEntityRepository repository;

@RequestMapping(method = RequestMethod.GET, value = "/{simpleEntityId}", produces =        MediaType.APPLICATION_JSON_VALUE)
public ResponseEntity<FileDatastore> getSimpleEntity(@PathVariable UUID simpleEntityId)     {
    SimpleEntity record = this.repository.findOneByUuid(simpleEntityId);
    HttpHeaders headers = new HttpHeaders();

    HttpStatus status = (record != null) ? HttpStatus.OK : HttpStatus.NOT_FOUND;

    return new ResponseEntity<>(record, headers, status);
}

Am I missing something?

Thanks for your help!

解决方案

Try annotate your UUID property with @org.hibernate.annotations.Type(type="org.hibernate.type.UUIDCharType")
or
@org.hibernate.annotations.Type(type="org.hibernate.type.UUIDBinaryType")

I faced a problem similar while query database with UUID due to MSB/LSB swap with UUID in binary format; we solved the problem treating data as String and do necessary conversion before conversion.

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