使用 Spring JPA 查询从序列中获取 nextval

Question

我有一个存储库，我试图用它从序列中获取下一个值。我需要参数化序列名称。

存储库查询看起来像：

@Query(value = "SELECT ?1.NEXTVAL from Dual;", nativeQuery = true) String getSeqID(@Param("seqName") String seqName);

但是，我收到以下异常：

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

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 2

在 PostgreSQL 中是：

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