Question

我有一个存储库，我正在尝试从序列中获取下一个值。我需要对序列名称进行参数化。

存储库查询类似于：

@Query（值=“从double;中选择SELECT？1.NEXTVAL”； nativeQuery = true）字符串getSeqID（@Param（“ seqName”）字符串seqName）;

但是，我遇到以下异常：

org.hibernate.QueryException: JPA-style positional param was not an integral ordinal
    at org.hibernate.engine.query.spi.ParameterParser.parse(ParameterParser.java:187) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]
    at org.hibernate.engine.query.spi.ParamLocationRecognizer.parseLocations(ParamLocationRecognizer.java:59) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]
    at org.hibernate.engine.query.internal.NativeQueryInterpreterStandardImpl.getParameterMetadata(NativeQueryInterpreterStandardImpl.java:34) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]
    at org.hibernate.engine.query.spi.QueryPlanCache.getSQLParameterMetadata(QueryPlanCache.java:125) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]

Answer 1

您可以使用EntityManager：

entityManager.createNativeQuery("select seqname.nextval ...").getSingleResult();

Answer 2

package br.com.mypackage.projectname.repository;

import java.math.BigDecimal;

import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.data.jpa.repository.Query;
import org.springframework.stereotype.Repository;

@Repository
public interface MyRepository extends JpaRepository<Myentity, Long> {

    @Query(value = "SELECT SCHEMA.SEQUENCE_NAME.nextval FROM dual", nativeQuery = true)
    public BigDecimal getNextValMySequence();

}

Answer 3

@Entity
@Table(name = "tabelName")
public class yourEntity{
    @Id
    @SequenceGenerator(name = "yourName", sequenceName = "yourSeqName", allocationSize = 1)
    @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "yourName")
    @Column(name = "id", updatable = false)
    protected Long id;
}

@Query(value = "SELECT yourSeqName.nextval FROM tableName", nativeQuery = true)
Long getNextSeriesId();

编辑：

Query q = em.createNativeQuery("SELECT yourSeqName.NEXTVAL FROM DUAL");
return (java.math.BigDecimal)q.getSingleResult();

Answer 4

在PostgreSQL中是：

Long value = Long.parseLong(entityManager
            .createNativeQuery("select nextval('sequence_name')")
            .getSingleResult().toString());

使用Spring JPA查询从序列中获取nextval

4 个答案: