Java/JPA with Error
[해결 방법] org.hibernate.QueryException
HJ0216
2023. 9. 20. 17:33
👉 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;
}
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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();
}
}
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🖨️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
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📡 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();
}
}
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⭐ JPQL은 엔티티 객체를 대상으로 쿼리를 작성하므로 col name이 아닌 field name으로 기재