[해결 방법] org.hibernate.QueryException

👉 기본 환경

- Language: Java

- DB: H2 Database

- IDE: IntelliJ

 

 

⌨️ 코드

Member Entity

 
@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

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

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

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