I'm using Spring Boot 1.5.2. I have a big Entity with multiple properties and multiple OneToMany relations, for example:
@Entity
@Table(name = "person")
class Person {
@Id
protected long id;
private String property1;
private String property2;
private String property3;
private String property4;
private String property5;
private String property6;
@OneToMany
private List<Obj1> obj1List;
@OneToMany
private List<Obj2> obj2List;
@OneToMany
private List<Obj3> obj3List;
@OneToMany
private List<Obj4> obj4List;
}
How can I read the list of Person
from database, but only with 2 properties id
and property2
, and update property2=0
.
Then, I can use JPA CrudRepository
to save():
public interface PersonRepository extends CrudRepository<Person, Long> {
}
for (Person person : personList) {
this.personRepository.save(person)
}
I don't want to use findAll()
from CrudRepository
which enables Hibernate to SELECT the whole list of Person with a big SQL query before saving to database.
CodePudding user response:
Inside your PersonRepository
interface, you should be able to add a Query
like this:
@Query("select new Person(id, property2) from Person")
List<Person> findIdAndProperty2();
The other fields should come back null as they haven't been specified in your query. You'll just need to add a constructor to Person
with id
and property2
as arguments.
For updating, you can use similar syntax...
@Modifying
@Query("update Person set property2 = ?1 where id in ?2")
int updateProperty2(String property2, List<Long> ids);
com.google.common.collect.Iterables.partition
can be used to process your updates in chunks. For instance...
for (List<Long> curUpdateIds : Iterables.partition(ids, 1000)) {
personRepository.updateProperty2("0", curUpdateIds);
}