Querydsl BooleanExpression 쿼리 최적화

Question

Querydsl BooleanExpressions를 반환하는 정적 메서드로 클래스를 만들었으므로이 스프링 표현식을 Spring 데이터 저장소의 findAll() 또는 findOne() 메서드에 전달하고 응용 프로그램 전체에서 다시 사용할 수 있습니다. 여기처럼 : http://www.petrikainulainen.net/programming/spring-framework/spring-data-jpa-tutorial-part-five-querydsl/.

내가 가지고있는 문제는 내 메서드에서 반환되는 BooleanExpression이 하위 최적화 된 것으로 보이는 쿼리를 생성하고 있다는 것입니다. 적어도 내 경우에는 그렇습니다.

public static BooleanExpression byExternalIdAndProviderId(
    final String externalId, final Long providerId) { 

    QEListing listing = QEListing.eListing; 

    return listing.externalIds.any().externalId.equalsIgnoreCase(externalId) 
     .and(listing.externalIds.any().provider.id.eq(providerId)); 
} 

쿼리 생성됩니다 다음 BooleanExpression을 반환

방법

이 같은 것을 반대로

select elisting0_.id as id1_24_, 
    elisting0_.address1 as address2_24_, 
    elisting0_.address2 as address3_24_, 
    elisting0_.business_name as business4_24_, 
    elisting0_.city_id as city10_24_, 
    elisting0_.created_date as created5_24_, 
    elisting0_.latitude as latitude6_24_, 
    elisting0_.longitude as longitud7_24_, 
    elisting0_.master_ext_id as master11_24_, 
    elisting0_.modified_date as modified8_24_, 
    elisting0_.state_id as state12_24_, 
    elisting0_.zip as zip13_24_, 
    elisting0_.zip_4 as zip9_24_, 
    elisting0_1_.vanity_name as vanity1_34_, 
    elisting0_2_.account_id as account1_2_ 
from listings elisting0_ 
    left outer join vanity_names elisting0_1_ on elisting0_.id=elisting0_1_.listing_id 
    left outer join accounts_to_listings elisting0_2_ on elisting0_.id=elisting0_2_.listing_id 
where (
    exists (
     select 1 
     from external_ids eexternali1_ 
     where (
      eexternali1_.id in (
       select externalid2_.external_id 
       from listing_to_external_id externalid2_ 
       where elisting0_.id=externalid2_.listing_id)) 
       and lower(eexternali1_.external_id)='123456' 
     ) 
    ) 
and (
    exists (
     select 1 
     from external_ids eexternali3_ 
     where (
      eexternali3_.id in (
       select externalid4_.external_id 
       from listing_to_external_id externalid4_ 
       where elisting0_.id=externalid4_.listing_id 
      ) 
     ) 
     and eexternali3_.provider_id=1 
    ) 
) 

:

select elisting0_.id as id1_24_, 
    elisting0_.address1 as address2_24_, 
    elisting0_.address2 as address3_24_, 
    elisting0_.business_name as business4_24_, 
    elisting0_.city_id as city10_24_, 
    elisting0_.created_date as created5_24_, 
    elisting0_.latitude as latitude6_24_, 
    elisting0_.longitude as longitud7_24_, 
    elisting0_.master_ext_id as master11_24_, 
    elisting0_.modified_date as modified8_24_, 
    elisting0_.state_id as state12_24_, 
    elisting0_.zip as zip13_24_, 
    elisting0_.zip_4 as zip9_24_, 
    elisting0_1_.vanity_name as vanity1_34_, 
    elisting0_2_.account_id as account1_2_ 
from listings elisting0_ 
    left outer join vanity_names elisting0_1_ on elisting0_.id=elisting0_1_.listing_id 
    left outer join accounts_to_listings elisting0_2_ on elisting0_.id=elisting0_2_.listing_id 
    join listing_to_external_id a on elisting0_.id = a.listing_id 
    join external_ids b on a.external_id = b.id 
where lower(b.external_id) = '123456' 
and b.provider_id = 1 

를 생성하는 방법이 있나요 더 최적의 쿼리를 제공하면서도 여전히 BooleanExpressions의 재사용 성을 제공하고 Spring D ata의 저장소?

Answer 1

결과 집합을 수정하기 때문에 더 최적의 방법 일 수 있지만 조인을 사용할 수 없습니다. 따라서 Querydsl에서 any() 서브 쿼리를 최적화하는 방법을 찾으면 GitHub 이슈 트래커에서 티켓을 생성하십시오.

Querydsl 사용이 술어로 제한되어 있으면 조인을 제어 할 수 없습니다.