👉 기본 환경

- Language: Java

- DB: H2 Database

- IDE: IntelliJ

 

 

⌨️ 코드

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public class Main {
    public static void main(String[] args) {
 
        EntityManagerFactory emf = Persistence.createEntityManagerFactory("hello");
 
        EntityManager em = emf.createEntityManager();
 
        EntityTransaction tx = em.getTransaction();
        tx.begin(); // transaction 시작 선언
 
        try {
            Member member1 = new Member();
            member1.setName("Member1");
            em.persist(member1);
 
            Member member2 = new Member();
            member2.setName("Member1");
            em.persist(member2);
 
            TypedQuery<Member> query = em.createQuery("select m from Member m where m.name='Member2'", Member.class);
            Member findMember = query.getSingleResult();
 
            tx.commit(); // transaction 종료 후 commit
        } catch (Exception e) {
            tx.rollback(); // 문제가 생길 경우, rollback 진행
            e.printStackTrace();
        } finally {
            em.close(); // tx에 문제가 생기더라도 em 반드시 종료
        }
 
        emf.close();
 
    }
}
 
 

 

 

🖨️오류

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javax.persistence.NoResultException: No entity found for query
    at org.hibernate.query.internal.AbstractProducedQuery.getSingleResult(AbstractProducedQuery.java:1583)
    at jpql.Main.main(Main.java:29)
 
 

 

 

📡 원인

getSingleResult()의 result가 0개

 

 

📰 해결 방법

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public class Main {
    public static void main(String[] args) {
 
        EntityManagerFactory emf = Persistence.createEntityManagerFactory("hello");
 
        EntityManager em = emf.createEntityManager();
 
        EntityTransaction tx = em.getTransaction();
        tx.begin(); // transaction 시작 선언
 
        try {
            Member member1 = new Member();
            member1.setName("Member1");
            em.persist(member1);
 
            Member member2 = new Member();
            member2.setName("Member1");
            em.persist(member2);
 
            TypedQuery<Member> query = em.createQuery("select m from Member m where m.name='Member2'", Member.class);
            List<Member> resultList = query.getResultList();
 
            tx.commit(); // transaction 종료 후 commit
        } catch (Exception e) {
            tx.rollback(); // 문제가 생길 경우, rollback 진행
            e.printStackTrace();
        } finally {
            em.close(); // tx에 문제가 생기더라도 em 반드시 종료
        }
 
        emf.close();
 
    }
}
 
 

Exception 방지를 위해 getResultList() 사용

→ 결과가 0개일 때, 빈 객체 반환

 

👉 기본 환경

- Language: Java

- DB: H2 Database

- IDE: IntelliJ

 

 

⌨️ 코드

Order Entity

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@Entity
public class Order {
 
    @Id @GeneratedValue
    private Long id;
    private int orderAmount;
    @Embedded
    private Address address;
    @ManyToOne
    @JoinColumn(name = "PRODUCT_ID")
    private Product product;
 
}
 
 

 

 

🖨️오류

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WARN: GenerationTarget encountered exception accepting command : Error executing DDL "
    alter table Order 
       add constraint FKiqsf200773nxdox2df2woemp5 
       foreign key (PRODUCT_ID) 
       references Product" via JDBC Statement
org.hibernate.tool.schema.spi.CommandAcceptanceException: Error executing DDL "
    alter table Order 
       add constraint FKiqsf200773nxdox2df2woemp5 
       foreign key (PRODUCT_ID) 
       references Product" via JDBC Statement
    at org.hibernate.tool.schema.internal.exec.GenerationTargetToDatabase.accept(GenerationTargetToDatabase.java:67)
    at org.hibernate.tool.schema.internal.SchemaCreatorImpl.applySqlString(SchemaCreatorImpl.java:439)
    at org.hibernate.tool.schema.internal.SchemaCreatorImpl.applySqlStrings(SchemaCreatorImpl.java:423)
    at org.hibernate.tool.schema.internal.SchemaCreatorImpl.createFromMetadata(SchemaCreatorImpl.java:374)
    at org.hibernate.tool.schema.internal.SchemaCreatorImpl.performCreation(SchemaCreatorImpl.java:166)
    at org.hibernate.tool.schema.internal.SchemaCreatorImpl.doCreation(SchemaCreatorImpl.java:135)
    at org.hibernate.tool.schema.internal.SchemaCreatorImpl.doCreation(SchemaCreatorImpl.java:121)
    at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.performDatabaseAction(SchemaManagementToolCoordinator.java:156)
    at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.process(SchemaManagementToolCoordinator.java:73)
    at org.hibernate.internal.SessionFactoryImpl.<init>(SessionFactoryImpl.java:314)
    at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:468)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:1249)
    at org.hibernate.jpa.HibernatePersistenceProvider.createEntityManagerFactory(HibernatePersistenceProvider.java:56)
    at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:79)
    at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:54)
    at jpql.Main.main(Main.java:11)
Caused by: org.h2.jdbc.JdbcSQLSyntaxErrorException: Syntax error in SQL statement "
    ALTER TABLE ORDER[*] 
       ADD CONSTRAINT FKIQSF200773NXDOX2DF2WOEMP5 
       FOREIGN KEY (PRODUCT_ID) 
       REFERENCES PRODUCT"; expected "identifier"; SQL statement:
 
    alter table Order 
       add constraint FKiqsf200773nxdox2df2woemp5 
       foreign key (PRODUCT_ID) 
       references Product [42001-200]
    at org.h2.message.DbException.getJdbcSQLException(DbException.java:453)
    at org.h2.message.DbException.getJdbcSQLException(DbException.java:429)
    at org.h2.message.DbException.getSyntaxError(DbException.java:243)
    at org.h2.command.Parser.readColumnIdentifier(Parser.java:4976)
    at org.h2.command.Parser.readIdentifierWithSchema(Parser.java:4925)
    at org.h2.command.Parser.readIdentifierWithSchema(Parser.java:4954)
    at org.h2.command.Parser.parseAlterTable(Parser.java:7683)
    at org.h2.command.Parser.parseAlter(Parser.java:6983)
    at org.h2.command.Parser.parsePrepared(Parser.java:887)
    at org.h2.command.Parser.parse(Parser.java:843)
    at org.h2.command.Parser.parse(Parser.java:815)
    at org.h2.command.Parser.prepareCommand(Parser.java:738)
    at org.h2.engine.Session.prepareLocal(Session.java:657)
    at org.h2.server.TcpServerThread.process(TcpServerThread.java:278)
    at org.h2.server.TcpServerThread.run(TcpServerThread.java:183)
    at java.base/java.lang.Thread.run(Thread.java:833)
 
    at org.h2.message.DbException.getJdbcSQLException(DbException.java:453)
    at org.h2.engine.SessionRemote.done(SessionRemote.java:611)
    at org.h2.command.CommandRemote.prepare(CommandRemote.java:85)
    at org.h2.command.CommandRemote.<init>(CommandRemote.java:51)
    at org.h2.engine.SessionRemote.prepareCommand(SessionRemote.java:481)
    at org.h2.jdbc.JdbcConnection.prepareCommand(JdbcConnection.java:1235)
    at org.h2.jdbc.JdbcStatement.executeInternal(JdbcStatement.java:212)
    at org.h2.jdbc.JdbcStatement.execute(JdbcStatement.java:201)
    at org.hibernate.tool.schema.internal.exec.GenerationTargetToDatabase.accept(GenerationTargetToDatabase.java:54)
    ... 15 more
 
 

 

 

📡 원인

Order는 SQL의 키워드 중 하나이기 때문에 테이블 이름으로 사용 없음

 

 

📰 해결 방법

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@Entity
@Table(name = "ORDERS")
public class Order {
 
    @Id @GeneratedValue
    private Long id;
    private int orderAmount;
    @Embedded
    private Address address;
    @ManyToOne
    @JoinColumn(name = "PRODUCT_ID")
    private Product product;
 
}
 
 

Table name 별도 지정

 

👉 기본 환경

- Language: Java

- DB: H2 Database

- IDE: IntelliJ

 

 

⌨️ 코드

Member Entity

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@Entity
public class Member extends BaseEntity {
 
    @Id @GeneratedValue
    @Column(name = "MEMBER_ID")
    private Long id;
 
    @Column(name = "USERNAME")
    private String name;
 
    @ManyToOne(fetch = FetchType.EAGER) // Member : Team = Many : One
    @JoinColumn(name = "TEAM_ID"// JoinColum = FK
    private Team team;
 
}
 
 

Main

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public class Main {
    public static void main(String[] args) {
 
        // 생략
 
        try {
 
            List<Member> memberList = em.createQuery("select m from Member m where m.username like 'kim%'", Member.class)
                    .getResultList();
 
            for(Member member : memberList){
                System.out.println("member: " + member);
            }
 
            tx.commit();
        } catch (Exception e) {
            tx.rollback();
            e.printStackTrace();
        } finally {
            em.close();
        }
 
        emf.close();
 
    }
 
}
 
 

 

 

🖨️오류

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INFO: HHH000490: Using JtaPlatform implementation: [org.hibernate.engine.transaction.jta.platform.internal.NoJtaPlatform]
java.lang.IllegalArgumentException: org.hibernate.QueryException: could not resolve property: username of: jpa_basic.Member [select m from jpa_basic.Member m where m.username like '%kim%']
    at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:138)
    at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:181)
    at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:188)
    at org.hibernate.internal.AbstractSharedSessionContract.createQuery(AbstractSharedSessionContract.java:725)
    at org.hibernate.internal.AbstractSharedSessionContract.createQuery(AbstractSharedSessionContract.java:816)
    at org.hibernate.internal.AbstractSessionImpl.createQuery(AbstractSessionImpl.java:23)
    at jpa_basic.Main.main(Main.java:568)
Caused by: org.hibernate.QueryException: could not resolve property: username of: jpa_basic.Member [select m from jpa_basic.Member m where m.username like '%kim%']
    at org.hibernate.QueryException.generateQueryException(QueryException.java:120)
    at org.hibernate.QueryException.wrapWithQueryString(QueryException.java:103)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:220)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:144)
    at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:113)
    at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:73)
    at org.hibernate.engine.query.spi.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:155)
    at org.hibernate.internal.AbstractSharedSessionContract.getQueryPlan(AbstractSharedSessionContract.java:604)
    at org.hibernate.internal.AbstractSharedSessionContract.createQuery(AbstractSharedSessionContract.java:716)
    ... 3 more
Caused by: org.hibernate.QueryException: could not resolve property: username of: jpa_basic.Member
    at org.hibernate.persister.entity.AbstractPropertyMapping.propertyException(AbstractPropertyMapping.java:73)
    at org.hibernate.persister.entity.AbstractPropertyMapping.toType(AbstractPropertyMapping.java:67)
    at org.hibernate.persister.entity.AbstractEntityPersister.toType(AbstractEntityPersister.java:2044)
    at org.hibernate.hql.internal.ast.tree.FromElementType.getPropertyType(FromElementType.java:412)
    at org.hibernate.hql.internal.ast.tree.FromElement.getPropertyType(FromElement.java:520)
    at org.hibernate.hql.internal.ast.tree.DotNode.getDataType(DotNode.java:693)
    at org.hibernate.hql.internal.ast.tree.DotNode.prepareLhs(DotNode.java:268)
    at org.hibernate.hql.internal.ast.tree.DotNode.resolve(DotNode.java:208)
    at org.hibernate.hql.internal.ast.HqlSqlWalker.resolve(HqlSqlWalker.java:1053)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.expr(HqlSqlBaseWalker.java:1303)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.exprOrSubquery(HqlSqlBaseWalker.java:4771)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.comparisonExpr(HqlSqlBaseWalker.java:4381)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.logicalExpr(HqlSqlBaseWalker.java:2161)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.whereClause(HqlSqlBaseWalker.java:827)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.query(HqlSqlBaseWalker.java:621)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectStatement(HqlSqlBaseWalker.java:325)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.statement(HqlSqlBaseWalker.java:273)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:272)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:192)
    ... 9 more
 
 

 

 

📡 원인

JPQL 작성 시, where 조건문에 col name인 username을 기재

 

 

📰 해결 방법

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public class Main {
    public static void main(String[] args) {
 
        // 생략
 
        try {
 
            List<Member> memberList = em.createQuery("select m from Member m where m.name like '%kim%'", Member.class)
                    .getResultList();
 
            for(Member member : memberList){
                System.out.println("member: " + member);
            }
 
            tx.commit();
        } catch (Exception e) {
            tx.rollback();
            e.printStackTrace();
        } finally {
            em.close();
        }
 
        emf.close();
 
    }
 
}
 
 

⭐ JPQL은 엔티티 객체를 대상으로 쿼리를 작성하므로 col name이 아닌 field name으로 기재

 

👉 기본 환경

- Language: Java

- DB: H2 Database

- IDE: IntelliJ

 

 

⌨️ 코드

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@Entity
public class MemberClctn {
 
    @Id
    @GeneratedValue
    private long id;
 
    private String username;
 
    // Address
    @Embedded
    private Address homeAddress;
 
    @ElementCollection
    @CollectionTable(name = "FAVORITE_FOOD",
            joinColumns = @JoinColumn(name = "id"))
    private Set<String> favoriteFoods = new HashSet<>();
 
    @OneToMany(cascade = CascadeType.ALL, orphanRemoval = true)
    @JoinColumn(name = "id")
    private List<AddressEntity> addressesHist = new ArrayList<>();
 
}
 
 

 

 

🖨️오류

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ERROR: Referential integrity constraint violation: "FKAXD8WBKC203QQ3EKY4KJIV7MO: PUBLIC.ADDRESS FOREIGN KEY(ID) REFERENCES PUBLIC.MEMBERCLCTN(ID) (3)"; SQL statement:
/* insert jpa_basic.AddressEntity */ insert into ADDRESS (city, street, zipcode, id) values (?, ?, ?, ?) [23506-200]
javax.persistence.PersistenceException: org.hibernate.exception.ConstraintViolationException: could not execute batch
    at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:154)
    at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:181)
    at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:188)
    at org.hibernate.internal.SessionImpl.doFlush(SessionImpl.java:1356)
    at org.hibernate.internal.SessionImpl.flush(SessionImpl.java:1339)
    at jpa_basic.Main.main(Main.java:526)
Caused by: org.hibernate.exception.ConstraintViolationException: could not execute batch
    at org.hibernate.exception.internal.SQLStateConversionDelegate.convert(SQLStateConversionDelegate.java:109)
    at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:42)
    at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:113)
    at org.hibernate.engine.jdbc.batch.internal.BatchingBatch.performExecution(BatchingBatch.java:129)
    at org.hibernate.engine.jdbc.batch.internal.BatchingBatch.doExecuteBatch(BatchingBatch.java:105)
    at org.hibernate.engine.jdbc.batch.internal.AbstractBatchImpl.execute(AbstractBatchImpl.java:148)
    at org.hibernate.engine.jdbc.internal.JdbcCoordinatorImpl.executeBatch(JdbcCoordinatorImpl.java:198)
    at org.hibernate.engine.spi.ActionQueue.executeActions(ActionQueue.java:633)
    at org.hibernate.engine.spi.ActionQueue.lambda$executeActions$1(ActionQueue.java:478)
    at java.base/java.util.LinkedHashMap.forEach(LinkedHashMap.java:684)
    at org.hibernate.engine.spi.ActionQueue.executeActions(ActionQueue.java:475)
    at org.hibernate.event.internal.AbstractFlushingEventListener.performExecutions(AbstractFlushingEventListener.java:348)
    at org.hibernate.event.internal.DefaultFlushEventListener.onFlush(DefaultFlushEventListener.java:40)
    at org.hibernate.event.service.internal.EventListenerGroupImpl.fireEventOnEachListener(EventListenerGroupImpl.java:102)
    at org.hibernate.internal.SessionImpl.doFlush(SessionImpl.java:1352)
    ... 2 more
Caused by: org.h2.jdbc.JdbcBatchUpdateException: Referential integrity constraint violation: "FKAXD8WBKC203QQ3EKY4KJIV7MO: PUBLIC.ADDRESS FOREIGN KEY(ID) REFERENCES PUBLIC.MEMBERCLCTN(ID) (2)"; SQL statement:
/* insert jpa_basic.AddressEntity */ insert into ADDRESS (city, street, zipcode, id) values (?, ?, ?, ?) [23506-200]
    at org.h2.jdbc.JdbcPreparedStatement.executeBatch(JdbcPreparedStatement.java:1235)
    at org.hibernate.engine.jdbc.batch.internal.BatchingBatch.performExecution(BatchingBatch.java:119)
    ... 13 more
 
 

 

 

📡 원인

AddressEntity와 일대다 매핑키 키 이름을 id로 부여하고자 함

= MemberClctn의 PK값을 AddressEntity의 FK로 사용

이 때, FK로 사용하고자 하는 컬럼을 AddressEntity의 id 필드로 지정하여 MemberClctn의 기본 키를 참조하는 외래 키로 간주

PK값이 중복된 값을 갖을 수 없으므로 무결성 제약 조건 위반 오류 발생

 

 

📰 해결 방법

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@Entity
public class MemberClctn {
 
    @Id
    @GeneratedValue
    private long id;
 
    private String username;
 
    // Address
    @Embedded
    private Address homeAddress;
 
    @ElementCollection
    @CollectionTable(name = "FAVORITE_FOOD",
            joinColumns = @JoinColumn(name = "id"))
    private Set<String> favoriteFoods = new HashSet<>();
 
    @OneToMany(cascade = CascadeType.ALL, orphanRemoval = true)
    @JoinColumn(name = "MCLCTN_NAME")
    private List<AddressEntity> addressesHist = new ArrayList<>();
 
}
 
 

FK값을 갖는 새로운 Col 생성

 

⭐ @JoinColumn()

- 현재 엔티티(애노테이션이 적용된 클래스)가 매핑할 테이블의 컬럼

 

👉 기본 환경

- Language: Java

- DB: H2 Database

- IDE: IntelliJ

 

 

⌨️ 코드

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@Entity
public class MemberUsingEm {
 
    @Id @GeneratedValue
    private long id;
 
    private String username;
 
    // Period
    @Embedded
    private Period workPeriod;
 
    // Address
    @Embedded
    private Address homeAddress;
    @Embedded
    private Address workAddress;
 
}
 
 

 

 

🖨️오류

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WARN: HHH000457: Joined inheritance hierarchy [jpa_basic.Item] defined explicit @DiscriminatorColumn.  Legacy Hibernate behavior was to ignore the @DiscriminatorColumn.  However, as part of issue HHH-6911 we now apply the explicit @DiscriminatorColumn.  If you would prefer the legacy behavior, enable the `hibernate.discriminator.ignore_explicit_for_joined` setting (hibernate.discriminator.ignore_explicit_for_joined=true)
Exception in thread "main" javax.persistence.PersistenceException: [PersistenceUnit: hello] Unable to build Hibernate SessionFactory
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.persistenceException(EntityManagerFactoryBuilderImpl.java:1326)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:1252)
    at org.hibernate.jpa.HibernatePersistenceProvider.createEntityManagerFactory(HibernatePersistenceProvider.java:56)
    at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:79)
    at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:54)
    at jpa_basic.Main.main(Main.java:14)
Caused by: org.hibernate.MappingException: Repeated column in mapping for entity: jpa_basic.MemberUsingEm column: city (should be mapped with insert="false" update="false")
    at org.hibernate.mapping.PersistentClass.checkColumnDuplication(PersistentClass.java:862)
    at org.hibernate.mapping.PersistentClass.checkPropertyColumnDuplication(PersistentClass.java:880)
    at org.hibernate.mapping.PersistentClass.checkPropertyColumnDuplication(PersistentClass.java:876)
    at org.hibernate.mapping.PersistentClass.checkColumnDuplication(PersistentClass.java:902)
    at org.hibernate.mapping.PersistentClass.validate(PersistentClass.java:634)
    at org.hibernate.mapping.RootClass.validate(RootClass.java:267)
    at org.hibernate.boot.internal.MetadataImpl.validate(MetadataImpl.java:351)
    at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:464)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:1249)
    ... 4 more
 
 

 

 

📡 원인

@Embedded 타입의 Address 객체가 homeAddress와 workAddress가 2개 존재

MemberUsingEm에서 이름이 같은 Address 컬럼들이 중복해서 생성

 

 

📰 해결 방법

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@Entity
public class MemberUsingEm {
 
    @Id @GeneratedValue
    private long id;
 
    private String username;
 
    // Period
    @Embedded
    private Period workPeriod;
 
    // Address
    @Embedded
    private Address homeAddress;
    @Embedded
    @AttributeOverrides({
            @AttributeOverride(name="city",
                    column=@Column(name = "WORK_CITY")),
            @AttributeOverride(name="street",
                    column=@Column(name = "WORK_STREET")),
            @AttributeOverride(name="zipcode",
                    column=@Column(name = "WORK_ZIPCODE"))
    })
    private Address workAddress;
 
}
 
 

@AttributeOverrides 사용하여 중복되지 않는 새로운 컬럼 name 부여

 

 

🖨️ 실행 결과

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Hibernate: 
    
    create table MemberUsingEm (
       id bigint not null,
        city varchar(255),
        street varchar(255),
        zipcode varchar(255),
        username varchar(255),
        WORK_CITY varchar(255),
        WORK_STREET varchar(255),
        WORK_ZIPCODE varchar(255),
        endDate timestamp,
        startDate timestamp,
        primary key (id)
    )