我编写了从DB获取数据的代码,但它也触发了一个意外的查询:
@SuppressWarnings("unchecked")
@Transactional
public List<Job> getAppliedPositionsById(Long userId) {
// String currentDate = SQLDateFormator.getCurrentDateInSqlFormat();
String strQuery = "from Job x left join x.applications a where a.applicant.id = :userId";
Query query = entityManager.createQuery(strQuery);
query.setParameter("userId", userId);
return query.getResultList();
}在return query.getResultList();上,它会触发两个查询。由于第二个查询,我得到了异常。
两个查询
冬眠:
select
job0_.id as id1_5_0_,
applicatio1_.id as id1_1_1_,
job0_.close_date as close_da2_5_0_,
job0_.committee_chair_id as committe6_5_0_,
job0_.description as descript3_5_0_,
job0_.publish_date as publish_4_5_0_,
job0_.title as title5_5_0_,
applicatio1_.applicant_id as applican6_1_1_,
applicatio1_.current_job_institution as current_2_1_1_,
applicatio1_.current_job_title as current_3_1_1_,
applicatio1_.current_job_year as current_4_1_1_,
applicatio1_.cv_id as cv_id7_1_1_,
applicatio1_.job_id as job_id8_1_1_,
applicatio1_.research_statement_id as research9_1_1_,
applicatio1_.submit_date as submit_d5_1_1_,
applicatio1_.teaching_statement_id as teachin10_1_1_
from
jobs job0_
left outer join
applications applicatio1_
on job0_.id=applicatio1_.job_id
where
applicatio1_.applicant_id=?冬眠:
select
user0_.id as id1_8_0_,
user0_.address as address2_8_0_,
user0_.email as email3_8_0_,
user0_.first_name as first_na4_8_0_,
user0_.last_name as last_nam5_8_0_,
user0_.password as password6_8_0_,
user0_.phone as phone7_8_0_
from
users user0_
where
user0_.id=?关于Users表的第二个查询完全没有必要。
Job实体
@Id
@GeneratedValue
private Long id;
private String title;
private String description;
@Column(name = "publish_date")
private Date publishDate;
@Column(name = "close_date")
private Date closeDate;
@ManyToOne
@JoinColumn(name = "committee_chair_id")
private User committeeChair;
@ManyToMany
@JoinTable(name = "job_committee_members",
joinColumns = @JoinColumn(name = "job_id") ,
inverseJoinColumns = @JoinColumn(name = "user_id") )
@OrderBy("lastName asc")
private List<User> committeeMembers;
@OneToMany(mappedBy = "job")
@OrderBy("date asc")
private List<Application> applications;
}Application实体:
@Id
@GeneratedValue
private Long id;
@ManyToOne
private Job job;
@ManyToOne
private User applicant;
@Column(name = "submit_date")
private Date submitDate;
@Column(name = "current_job_title")
private String currentJobTitle;
@Column(name = "current_job_institution")
private String currentJobInstitution;
@Column(name = "current_job_year")
private Integer currentJobYear;
@ElementCollection
@CollectionTable(name = "application_degrees",
joinColumns = @JoinColumn(name = "application_id") )
@OrderBy("year desc")
private List<Degree> degrees;
@OneToOne
private File cv;
@OneToOne
@JoinColumn(name = "research_statement_id")
private File researchStatement;
@OneToOne
@JoinColumn(name = "teaching_statement_id")
private File teachingStatement;
@OneToMany(mappedBy = "application",
cascade = { CascadeType.MERGE, CascadeType.PERSIST })
@OrderColumn(name = "round_index")
private List<Round> rounds;
}User实体:
@Id
@GeneratedValue
private Long id;
@Column(unique = true, nullable = false)
private String email;
@Column(nullable = false)
private String password;
@Column(name = "first_name")
private String firstName;
@Column(name = "last_name")
private String lastName;
private String address;
private String phone;
@OneToMany(mappedBy = "applicant")
@OrderBy("id desc")
private List<Application> applications;
}发布于 2015-07-25 11:47:22
在JPA2.0规范中,默认设置如下:
OneToMany:懒惰 ManyToOne:热切 ManyToMany:懒惰 OneToOne:热切
在应用程序类中
@ManyToOne
private User applicant;如果你把它换成懒惰的话
@ManyToOne(fetch = FetchType.LAZY)它应该按你想的方式工作。
发布于 2015-07-25 15:25:03
正如luksch所指出的,您的模型为用户定义了一个@ManyToOne关系,默认情况下,每次加载作业实例(或模型中的应用程序实例)时,都会急切地获取。然而,切换它FetchType.LAZY现在可能会产生预期的结果。使用oneToOne或manyToOne,hibernate将不得不进行额外的查询,即使它很懒。只有当您指定关系的optional=false属性和FetchType.LAZY时,它才会自动将代理对象设置为用户属性值。这是因为Hibernate在检查DB之前无法知道用户属性是否存在或为null。根据您的模型,一个更合适的解决方案是更新查询,以便在一个查询中获取用户对象,如下所示:
String strQuery = "from Job x left join fetch x.committeeChair u left join x.applications a where a.applicant.id = :userId"重要的部分是左连接获取x.committeeChair u,它告诉hibernate添加一个额外的连接并获取相关的对象。
这修复了使用JPQL获取作业实例时的行为。如果尝试通过id通过EntityManager.find方法加载单个作业实例。它仍然会为用户committeeChair生成一个额外的查询。您的加载策略可以通过使用Hibernate特定的(还不是JPA标准)获取模式进一步优化--请注意,获取模式可能会禁用可能不需要的延迟加载。我建议您首先决定需要加载哪些数据,始终存在哪些数据,并在此基础上优化查询。通常,额外的查询比在一个查询中加载整个实体图更好。祝好运
https://stackoverflow.com/questions/31625845
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