인프런 커뮤니티 질문&답변

leemsdong2

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실전! 스프링 데이터 JPA

Projections

findProjectionsByUsername 생성오류

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Error creating bean with name 'memberRepository' defined in study.datajpa.repository.MemberRepository defined in @EnableJpaRepositories declared on JpaRepositoriesRegistrar.EnableJpaRepositoriesConfiguration: Invocation of init method failed; nested exception is org.springframework.data.repository.query.QueryCreationException: Could not create query for public abstract java.util.List study.datajpa.repository.MemberRepository.findProjectionsByUsername(java.lang.String)! Reason: Failed to create query for method public abstract java.util.List study.datajpa.repository.MemberRepository.findProjectionsByUsername(java.lang.String)! null; nested exception is java.lang.IllegalArgumentException: Failed to create query for method public abstract java.util.List study.datajpa.repository.MemberRepository.findProjectionsByUsername(java.lang.String)! null

이렇게 repository에서 저 네임메서드가 생성이 안되어서 오류가 뜨는데 해결이 안됩니다..

java spring spring-boot JPA

퀴즈

Spring Data JPA에서 Projections를 사용하는 주된 목적은 무엇인가요?

복잡한 동적 검색 조건 구성

엔티티의 특정 필드만 선택하여 조회

원시 SQL 쿼리 직접 실행

예제 객체로 간단히 검색

답변 2

박형근

저도 부트 2.6.3에서는 안되는거같습니다.

2.6.2로 낮추니까 돌아가네요

David

안녕하세요. leemsdong2님, 공식 서포터즈 David입니다.

해당 리포지토리 코드도 함께 올려주세요.

감사합니다.

leemsdong2

질문자

public interface MemberRepository extends JpaRepository<Member,Long>, MemberRepositoryCustom{


    @Query("select m from Member m where m.username=:name and m.age =:age")
    List<Member> findUser(@Param("name") String username, @Param("age") int age);

    @Query("select m.username from Member m")
    List<String> findUsernameList();

    @Query("select new study.datajpa.dto.MemberDto(m.id, m.username, t.name) from Member m join m.team t")
    List<MemberDto> findMemberDto();

    @Query("select m from Member m where m.username in :names")
    List<Member> findByNames(@Param("names") List<String> names);

//    @Query("select m from Member m",
//            countQuery = "select count(m.username) from Member m")
    Page<Member> findByAge(int age, Pageable pageable);

    @Modifying(clearAutomatically = true)
    @Query("update Member m set m.age=m.age+1 where m.age>=:age")
    int bulkAgePlus(@Param("age") int age);

    @Query("select m from Member m join fetch m.team")
    List<Member> findMemberFetchJoin();

    @EntityGraph(attributePaths = {"team"})
    List<Member> findByAge(@Param("age") int age);

    List<UsernameOnly> findProjectionsByUsername(String username);

}

leemsdong2

질문자

public interface UsernameOnly {

    String getUsername();
}

혹시해서 이것도 올립니다

David

Member 클래스도 올려주실 수 있으세요?

leemsdong2

질문자

@Entity
@Getter
@Setter
@NoArgsConstructor(access = AccessLevel.PROTECTED)
@ToString(of = {"id","username","team"})
public class Member extends BaseEntity{

    @Id @GeneratedValue
    @Column(name = "member_id")
    private Long id;

    private String username;

    private int age;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "team_id")
    private Team team;

    public Member(String username) {
        this.username = username;
    }

    public Member(String username, int i, Team team) {
        this.username = username;
        this.age = i;
        if(team!=null){
            changeTeam(team);
        }
    }

    public Member(String username, int age) {
        this(username, age, null);
    }


    public void changeTeam(Team team){
        this.team = team;
        team.getMembers().add(this);
    }
}

David

혹시 Spring Data Jpa 버전이 최신버전이실까요?

만약 최신버전이시라면 아래 글 답변을 참고하셔서 Spring Boot 버전 또는 Spring Data Jpa 버전을 낮춰보시고 다시 시도해주시겠어요?

https://www.inflearn.com/questions/444705

leemsdong2

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