问题
I'm working on a migration software that will consume unknown data from REST services.
I already think about use MongoDB but I decide to not use it and use PostgreSQL.
After read this I'm trying to implement it in my SpringBoot app using Spring JPA but I don't know to map jsonb
in my entity.
Tried this but understood nothing!
Here is where I am:
@Repository
@Transactional
public interface DnitRepository extends JpaRepository<Dnit, Long> {
@Query(value = "insert into dnit(id,data) VALUES (:id,:data)", nativeQuery = true)
void insertdata( @Param("id")Integer id,@Param("data") String data );
}
and ...
@RestController
public class TestController {
@Autowired
DnitRepository dnitRepository;
@RequestMapping(value = "/dnit", method = RequestMethod.GET)
public String testBig() {
dnitRepository.insertdata(2, someJsonDataAsString );
}
}
and the table:
CREATE TABLE public.dnit
(
id integer NOT NULL,
data jsonb,
CONSTRAINT dnit_pkey PRIMARY KEY (id)
)
How can I do this?
Note: I don't want/need an Entity to work on. My JSON will always be String but I need jsonb to query the DB
回答1:
You are making things overly complex by adding Spring Data JPA just to execute a simple insert statement. You aren't using any of the JPA features. Instead do the following
- Replace
spring-boot-starter-data-jpa
withspring-boot-starter-jdbc
- Remove your
DnitRepository
interface - Inject
JdbcTemplate
where you where injectingDnitRepository
- Replace
dnitRepository.insertdata(2, someJsonDataAsString );
withjdbcTemplate.executeUpdate("insert into dnit(id, data) VALUES (?,to_json(?))", id, data);
You were already using plain SQL (in a very convoluted way), if you need plain SQL (and don't have need for JPA) then just use SQL.
Ofcourse instead of directly injecting the JdbcTemplate
into your controller you probably want to hide that logic/complexity in a repository or service.
回答2:
Tried this but understood nothing!
To fully work with jsonb
in Spring Data JPA (Hibernate) project with Vlad Mihalcea's hibernate-types lib you should just do the following:
1) Add this lib to your project:
<dependency>
<groupId>com.vladmihalcea</groupId>
<artifactId>hibernate-types-52</artifactId>
<version>2.2.2</version>
</dependency>
2) Then use its types in your entities, for example:
@Data
@NoArgsConstructor
@Entity
@Table(name = "parents")
@TypeDef(name = "jsonb", typeClass = JsonBinaryType.class)
public class Parent implements Serializable {
@Id
@GeneratedValue(strategy = SEQUENCE)
private Integer id;
@Column(length = 32, nullable = false)
private String name;
@Type(type = "jsonb")
@Column(columnDefinition = "jsonb")
private List<Child> children;
@Type(type = "jsonb")
@Column(columnDefinition = "jsonb")
private Bio bio;
public Parent(String name, List children, Bio bio) {
this.name = name;
this.children = children;
this.bio = bio;
}
}
@Data
@NoArgsConstructor
@AllArgsConstructor
public class Child implements Serializable {
private String name;
}
@Data
@NoArgsConstructor
@AllArgsConstructor
public class Bio implements Serializable {
private String text;
}
Then you will be able to use, for example, a simple JpaRepository
to work with your objects:
public interface ParentRepo extends JpaRepository<Parent, Integer> {
}
parentRepo.save(new Parent(
"parent1",
asList(new Child("child1"), new Child("child2")),
new Bio("bio1")
)
);
Parent result = parentRepo.findById(1);
List<Child> children = result.getChildren();
Bio bio = result.getBio();
来源:https://stackoverflow.com/questions/51276703/how-to-store-postgresql-jsonb-using-springboot-jpa