如何调整休眠查询

问题描述

| 我不得不调试一些我未编写的休眠代码。我会直截了当地承认我正在努力使Hibernate如何将查询整合到这一点，所以请耐心等待。 我们看到的问题是，Web应用程序花费大量时间来加载页面（使用JSP，Spring和Hibernate）。 正在执行的查询之一引用了一个映射中的公式，该公式似乎在休眠执行的查询中使用了两次。第一个引用是列值，第二个引用用作内部联接的条件。第二个是内部联接，它强制执行我们不需要的全表扫描，更不用说它使查询真正变慢了。 有没有一种方法可以调整休眠查询，以使其不在查询中使用此公式（列值除外）？请告知您是否需要其他信息。 编辑：根据要求，这是我从记录休眠状态中获得的查询：

SELECT this_.PARTY_ID       AS PARTY1_23_2_,this_.STUDENT_ID          AS STUDENT3_23_2_,this_.STUDENT_PIDM        AS STUDENT4_23_2_,this_.USERNAME            AS USERNAME23_2_,this_.FirsT_NAME          AS FirsT6_23_2_,this_.LAST_NAME           AS LAST7_23_2_,this_.ACTIVE              AS ACTIVE23_2_,subjectenr1_.PARTY_ID     AS PARTY1_24_0_,subjectenr1_.OFFERING_ID  AS OFFERING2_24_0_,subjectenr1_.RSTS_CODE    AS RSTS3_24_0_,subjectenr1_.SITE_CODE    AS SITE4_24_0_,subjectenr1_.PROGRAM_CODE AS PROGRAM5_24_0_,(SELECT su2.offering_id
  FROM sakaicfg.subject_offering su2
  WHERE su2.offering_id = subjectenr1_.OFFERING_ID
  )                               AS formula0_0_,subjectoff2_.OFFERING_ID        AS OFFERING1_25_1_,subjectoff2_.CRN                AS CRN25_1_,subjectoff2_.IS_ACTIVE          AS IS3_25_1_,subjectoff2_.TEACHING_SCHOOL    AS TEACHING4_25_1_,subjectoff2_.CAmpuS             AS CAmpuS25_1_,subjectoff2_.START_SEMESTER     AS START6_25_1_,subjectoff2_.STUDYMODE          AS STUDYMODE25_1_,subjectoff2_.SUBJECT_ID         AS SUBJECT8_25_1_,subjectoff2_.GRADE_MARKING_CODE AS GRADE9_25_1_,(SELECT MAX(su2.effective_semester)
  FROM sakaicfg.subject su2
  WHERE su2.subject_id        = subjectoff2_.SUBJECT_ID
  AND su2.effective_semester <= subjectoff2_.START_SEMESTER
  ) AS formula1_1_
FROM SAKAICFG.STUDENT this_
INNER JOIN SAKAICFG.SUBJECT_ENROLMENT subjectenr1_
ON this_.PARTY_ID=subjectenr1_.PARTY_ID
INNER JOIN SAKAICFG.SUBJECT_OFFERING subjectoff2_
ON (SELECT su2.offering_id
  FROM sakaicfg.subject_offering su2
  WHERE su2.offering_id     = subjectenr1_.OFFERING_ID)=subjectoff2_.OFFERING_ID
WHERE this_.ACTIVE          =\'Y\'
AND subjectenr1_.RSTS_CODE <>\'DD\'
AND subjectoff2_.IS_ACTIVE  =\'Y\'
AND subjectoff2_.OFFERING_ID=\'35505\'
ORDER BY this_.PARTY_ID DESC;

有问题的部分是以下摘录：

  (SELECT su2.offering_id
  FROM sakaicfg.subject_offering su2
  WHERE su2.offering_id = subjectenr1_.OFFERING_ID
  )

“主题注册”表的映射xml文件包含以下关系：

<many-to-one name=\"student\" entity-name=\"CsuActiveStudentDto\" column=\"PARTY_ID\" insert=\"false\" update=\"false\" access=\"field\" unique=\"false\"/>
<many-to-one name=\"subjectOffering\" entity-name=\"CsuSubjectOfferingDto\" insert=\"false\" update=\"false\" access=\"field\" unique=\"false\">
   <formula>(select su2.offering_id
            from sakaicfg.subject_offering su2 where su2.offering_id = OFFERING_ID)</formula>
</many-to-one>
<one-to-one name=\"student\" class=\"au.edu.csu.enterprise.domain.StudentDto\" property-ref=\"enrolment\"/>
<many-to-one name=\"offering\" class=\"au.edu.csu.enterprise.domain.SubjectOfferingDto\" column=\"OFFERING_ID\"/>

我一直在与我们的一名DBA一起研究查询，并隔离那些需要研究的查询，这就是我们找到该查询的方式。     

解决方法

您可以通过自定义SQL（而非hql）覆盖每个实体（以及集合）的Hibernate Create Read Update和Delete语句。 有关更多详细信息，请参见《 Hibernate参考：第17.3章》。在Hibernate Annotation Reference：第2.4.10章中，用于创建，更新和删除以及其工作方式的自定义SQL。用于CRUD操作的自定义SQL

@Entity
@SQLInsert( sql=\"INSERT INTO Demo(name,id) VALUES(?,?)\")
@Loader(namedQuery = \"betterLoad\")
@NamedNativeQuery(name=\"betterLoad\",query=\"select id,name from Demo where id= ?\",resultClass = Demo.class)
public class Demo {
    @Id
    private Long id;
    private String name;
}