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我已经按照我的仓库本地查询方法:春数据JPA,原生查询,返回错误的字段类型
@Query(value="SELECT appSub.ApplicationFormId as appFormId, appSub.profileId as profileId, "
+ "p.CASId as profileCASId, ps.programId as programId FROM [unicas_config].[dbo].ApplicationFormEarlyDecisionConfig appFormED "
+ "INNER JOIN [unicas_ux].[dbo].ApplicationSubmission appSub ON appFormED.ApplicationFormId = appSub.applicationFormId "
+ "INNER JOIN [unicas_ux].[dbo].Profile p ON appSub.profileId = p.id "
+ "INNER JOIN [unicas_ux].[dbo].ProgramSelected ps ON p.id=ps.ProfileId AND appSub.applicationFormId = ps.instanceId "
+ "WHERE appFormED.EarlyDecisionVerdictDate >=:fromDate AND appFormED.EarlyDecisionVerdictDate <:toDate "
+ "AND appSub.EarlyDecisionStatus='Applied Early Decision' "
+ "AND appSub.ApplicationStatus='Received' "
+ "AND ps.IsPaid =1 "
+ "ORDER BY appSub.ApplicationFormId",nativeQuery = true)
List<Object[]> getAllEarlyDecisionApplicantsWithPaidProgramsOnVerdictDate(@Param("fromDate") Date fromDate, @Param("toDate") Date toDate);
现在,我要地图返回结果:
long appFormId = (Long)obj[0]
long profileId = (Long)obj[1]
long programId = (Long)obj[3]
当我我正在这样做,我得到java.lang.ClassCastException: java.lang.Integer cannot be cast to java.lang.Long
作为Hibernate考虑Integer
类型的ID代替Long
。
请告诉我如何以编程方式告诉Hibernate返回正确的类型。
我不认为这与Hibernate有很大关系。 JDBC驱动程序将返回与数据库中列类型对应的Java类型。为什么你不能只使用Java Integer? –
@AlanHay。是的,你赖特。我发现DB中的列类型是Int类型,即使实体ID也是长类型的。这就是为什么我得到Integer而不是很长。 –