Question

我已经调试了一些我没写的hibernate代码。我会直截了当地承认，我很难理解Hibernate如何将查询汇总到这一点，所以请耐心等待。

我们看到的问题是Web应用程序花费了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似乎在hibernate执行的查询中使用了两次的映射的公式。第一个引用是列值，第二个引用用作内连接的条件。内连接中的第二个是强制执行我们不需要的全表扫描，更不用说它使查询真的很慢。

有没有办法调整hibernate查询，以便它不在查询中使用此公式（列值除外）？如果您需要更多信息，请告知。

编辑：根据要求，这是我从日志记录中获取的查询：

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;

有问题的部分是摘录：

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

Subject Enrollment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DBA一起工作，查看查询并隔离需要查看的查询，这就是我们找到这个查询的方式。

Answer 1

您可以通过自定义SQL（而不是hql）为每个实体（以及集合）覆盖Hibernate Create Read Update和Delete语句。

有关更多详细信息，请参阅Hibernate Reference：Chapter 17.3. Custom SQL for create, update and delete以及它在Hibernate Annotation Reference中的工作原理：Chapter 2.4.10. Custom SQL for CRUD operations

@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;
}

如何调整Hibernate查询

1 个答案: