Я использую Hibernate и Spring для создания Rest API. Я сопоставляю Trainer с таблицей в mySQL
Caused by: java.lang.IllegalArgumentException: Validation failed for query for method public abstract java.util.List io.fruitful.model.repos.TrainerRepository.getTrainerByFilter(java.lang.String,java.lang.Integer)!
at org.springframework.data.jpa.repository.query.SimpleJpaQuery.validateQuery(SimpleJpaQuery.java:93) ~[spring-data-jpa-2.0.8.RELEASE.jar:2.0.8.RELEASE]
at org.springframework.data.jpa.repository.query.SimpleJpaQuery.<init>(SimpleJpaQuery.java:63) ~[spring-data-jpa-2.0.8.RELEASE.jar:2.0.8.RELEASE]
at org.springframework.data.jpa.repository.query.JpaQueryFactory.fromMethodWithQueryString(JpaQueryFactory.java:76) ~[spring-data-jpa-2.0.8.RELEASE.jar:2.0.8.RELEASE]
at org.springframework.data.jpa.repository.query.JpaQueryFactory.fromQueryAnnotation(JpaQueryFactory.java:56) ~[spring-data-jpa-2.0.8.RELEASE.jar:2.0.8.RELEASE]
at org.springframework.data.jpa.repository.query.JpaQueryLookupStrategy$DeclaredQueryLookupStrategy.resolveQuery(JpaQueryLookupStrategy.java:139) ~[spring-data-jpa-2.0.8.RELEASE.jar:2.0.8.RELEASE]
at org.springframework.data.jpa.repository.query.JpaQueryLookupStrategy$CreateIfNotFoundQueryLookupStrategy.resolveQuery(JpaQueryLookupStrategy.java:206) ~[spring-data-jpa-2.0.8.RELEASE.jar:2.0.8.RELEASE]
at org.springframework.data.jpa.repository.query.JpaQueryLookupStrategy$AbstractQueryLookupStrategy.resolveQuery(JpaQueryLookupStrategy.java:79) ~[spring-data-jpa-2.0.8.RELEASE.jar:2.0.8.RELEASE]
at org.springframework.data.repository.core.support.RepositoryFactorySupport$QueryExecutorMethodInterceptor.lookupQuery(RepositoryFactorySupport.java:553) ~[spring-data-commons-2.0.8.RELEASE.jar:2.0.8.RELEASE]
at org.springframework.data.repository.core.support.RepositoryFactorySupport$QueryExecutorMethodInterceptor.lambda$mapMethodsToQuery$1(RepositoryFactorySupport.java:546) ~[spring-data-commons-2.0.8.RELEASE.jar:2.0.8.RELEASE]
at java.util.stream.ReferencePipeline$3$1.accept(ReferencePipeline.java:193) ~[na:1.8.0_144]
at java.util.Iterator.forEachRemaining(Iterator.java:116) ~[na:1.8.0_144]
at java.util.Collections$UnmodifiableCollection$1.forEachRemaining(Collections.java:1049) ~[na:1.8.0_144]
at java.util.Spliterators$IteratorSpliterator.forEachRemaining(Spliterators.java:1801) ~[na:1.8.0_144]
at java.util.stream.AbstractPipeline.copyInto(AbstractPipeline.java:481) ~[na:1.8.0_144]
at java.util.stream.AbstractPipeline.wrapAndCopyInto(AbstractPipeline.java:471) ~[na:1.8.0_144]
at java.util.stream.ReduceOps$ReduceOp.evaluateSequential(ReduceOps.java:708) ~[na:1.8.0_144]
at java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:234) ~[na:1.8.0_144]
at java.util.stream.ReferencePipeline.collect(ReferencePipeline.java:499) ~[na:1.8.0_144]
at org.springframework.data.repository.core.support.RepositoryFactorySupport$QueryExecutorMethodInterceptor.mapMethodsToQuery(RepositoryFactorySupport.java:548) ~[spring-data-commons-2.0.8.RELEASE.jar:2.0.8.RELEASE]
at org.springframework.data.repository.core.support.RepositoryFactorySupport$QueryExecutorMethodInterceptor.lambda$new$0(RepositoryFactorySupport.java:538) ~[spring-data-commons-2.0.8.RELEASE.jar:2.0.8.RELEASE]
at java.util.Optional.map(Optional.java:215) ~[na:1.8.0_144]
и
Caused by: org.hibernate.hql.internal.ast.QuerySyntaxException: trainers is not mapped [SELECT t from trainers t WHERE t.trainerName = :name AND t.accountId = :id]
at org.hibernate.hql.internal.ast.QuerySyntaxException.generateQueryException(QuerySyntaxException.java:79) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.hibernate.QueryException.wrapWithQueryString(QueryException.java:103) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:217) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.hibernate.hql.internal.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:141) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:115) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:77) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.hibernate.engine.query.spi.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:153) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.hibernate.internal.AbstractSharedSessionContract.getQueryPlan(AbstractSharedSessionContract.java:553) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
at org.hibernate.internal.AbstractSharedSessionContract.createQuery(AbstractSharedSessionContract.java:662) ~[hibernate-core-5.2.17.Final.jar:5.2.17.Final]
... 79 common frames omitted
Организация:
@Entity
@Table(name = "trainers")
public class Trainer implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "trainer_id", unique = true, nullable = false)
private Integer trainerId;
@Column (name = "trainer_name")
private String trainerName;
@Column (name = "account_id")
private Integer accountId;
Вот код репо:
@Query(value = "SELECT t from trainers t WHERE t.trainerName = :name AND t.accountId = :id")
List<Trainer> getTrainerByFilter(@Param("name") String name,
@Param("id") Integer id);
@Query (значение = «ВЫБЕРИТЕ * из тренеров, где account_id имеет значение null», nativeQuery = true) Список getTrainerNoAccount();
Конфигурационный SQL:
environment.setPackagesToScan("io.fruitful.model");
получитьTrainerNoAccount() работает нормально.
Hibernate отображает вашу сущность как Trainer, имя вашего класса, поскольку вы не указали параметр имени в аннотации @Entity.
Таким образом, ваш запрос должен быть "SELECT t from Тренер t...".
getTrainerNoAccount() работает нормально, потому что его @Query установлен как собственный, используя имя таблицы, а не имя вашей сущности.
@Query(value = "SELECT t from Trainer t WHERE t.trainerName = :name AND t.accountId = :id")
List<Trainer> getTrainerByFilter(@Param("name") String name,
@Param("id") Integer id);
Измените свой запрос таким образом, потому что JPA требуется имя компонента, а не имя таблицы.
читайте документацию JPA. JPQL использует имена сущностей. это не SQL