为什么hibernate在id上搜索时使用JOIN，而不是在另一个字段上？

Question

我目前使用Spring Data和存储库来获取我的数据。当使用默认值时：

myRepo.findById(id)

生成一个使用联接的查询：

Hibernate: select questionna0_.id as id1_6_0_, questionna0_.action as action2_6_0_, questionna0_.first_question_id as first_qu3_6_0_, answers1_.questionnaire_id as question5_0_1_, answers1_.id as id1_0_1_, answers1_.id as id1_0_2_, answers1_.order as order2_0_2_, answers1_.question_id as question3_0_2_, answers1_.value as value4_0_2_, validation2_.answer_id as answer_i2_1_3_, validation2_.id as id1_1_3_, validation2_.id as id1_1_4_, validation2_.answer_id as answer_i2_1_4_, validation2_.operator as operator3_1_4_, validation2_.value as value4_1_4_, questions3_.questionnaire_id as question7_5_5_, questions3_.id as id1_5_5_, questions3_.id as id1_5_6_, questions3_.answer_type as answer_t2_5_6_, questions3_.optional as optional3_5_6_, questions3_.question_key as question4_5_6_, questions3_.question_label as question5_5_6_, questions3_.question_type as question6_5_6_, answers4_.question_id as question3_0_7_, answers4_.id as id1_0_7_, answers4_.id as id1_0_8_, answers4_.order as order2_0_8_, answers4_.question_id as question3_0_8_, answers4_.value as value4_0_8_, branches5_.question_id as question3_2_9_, branches5_.id as id1_2_9_, branches5_.id as id1_2_10_, branches5_.question_id_target as question2_2_10_, conditions6_.branch_id as branch_i5_3_11_, conditions6_.id as id1_3_11_, conditions6_.id as id1_3_12_, conditions6_.operator as operator2_3_12_, conditions6_.question_id as question3_3_12_, conditions6_.value as value4_3_12_ from questionnaire questionna0_ left outer join answer answers1_ on questionna0_.id=answers1_.questionnaire_id left outer join answer_validation validation2_ on answers1_.id=validation2_.answer_id left outer join question questions3_ on questionna0_.id=questions3_.questionnaire_id left outer join answer answers4_ on questions3_.id=answers4_.question_id left outer join branch branches5_ on questions3_.id=branches5_.question_id left outer join branch_condition conditions6_ on branches5_.id=conditions6_.branch_id where questionna0_.id=?

但是，当使用同一个存储库对另一个字段进行查询时，要执行几乎相同的操作：

myRepo.findByAction(action)

Hibernate: select questionna0_.id as id1_6_, questionna0_.action as action2_6_, questionna0_.first_question_id as first_qu3_6_ from questionnaire questionna0_ where questionna0_.action=?
Hibernate: select questions0_.questionnaire_id as question7_5_0_, questions0_.id as id1_5_0_, questions0_.id as id1_5_1_, questions0_.answer_type as answer_t2_5_1_, questions0_.optional as optional3_5_1_, questions0_.question_key as question4_5_1_, questions0_.question_label as question5_5_1_, questions0_.question_type as question6_5_1_ from question questions0_ where questions0_.questionnaire_id=?
Hibernate: select answers0_.question_id as question3_0_0_, answers0_.id as id1_0_0_, answers0_.id as id1_0_1_, answers0_.order as order2_0_1_, answers0_.question_id as question3_0_1_, answers0_.value as value4_0_1_ from answer answers0_ where answers0_.question_id=?
Hibernate: select validation0_.answer_id as answer_i2_1_0_, validation0_.id as id1_1_0_, validation0_.id as id1_1_1_, validation0_.answer_id as answer_i2_1_1_, validation0_.operator as operator3_1_1_, validation0_.value as value4_1_1_ from answer_validation validation0_ where validation0_.answer_id=?
Hibernate: select branches0_.question_id as question3_2_0_, branches0_.id as id1_2_0_, branches0_.id as id1_2_1_, branches0_.question_id_target as question2_2_1_ from branch branches0_ where branches0_.question_id=?
Hibernate: select conditions0_.branch_id as branch_i5_3_0_, conditions0_.id as id1_3_0_, conditions0_.id as id1_3_1_, conditions0_.operator as operator2_3_1_, conditions0_.question_id as question3_3_1_, conditions0_.value as value4_3_1_ from branch_condition conditions0_ where conditions0_.branch_id=?
Hibernate: select answers0_.question_id as question3_0_0_, answers0_.id as id1_0_0_, answers0_.id as id1_0_1_, answers0_.order as order2_0_1_, answers0_.question_id as question3_0_1_, answers0_.value as value4_0_1_ from answer answers0_ where answers0_.question_id=?
Hibernate: select validation0_.answer_id as answer_i2_1_0_, validation0_.id as id1_1_0_, validation0_.id as id1_1_1_, validation0_.answer_id as answer_i2_1_1_, validation0_.operator as operator3_1_1_, validation0_.value as value4_1_1_ from answer_validation validation0_ where validation0_.answer_id=?
Hibernate: select branches0_.question_id as question3_2_0_, branches0_.id as id1_2_0_, branches0_.id as id1_2_1_, branches0_.question_id_target as question2_2_1_ from branch branches0_ where branches0_.question_id=?
Hibernate: select answers0_.questionnaire_id as question5_0_0_, answers0_.id as id1_0_0_, answers0_.id as id1_0_1_, answers0_.order as order2_0_1_, answers0_.question_id as question3_0_1_, answers0_.value as value4_0_1_ from answer answers0_ where answers0_.questionnaire_id=?

@Entity
@Table(name = "questionnaire")
data class QuestionnaireEntity(
    @Id
    @GeneratedValue(generator = "UUID")
    @GenericGenerator(name = "UUID", strategy = "org.hibernate.id.UUIDGenerator")
    val id: UUID?,
    val action: String,
    val firstQuestionId: UUID?,
    @OneToMany(fetch = FetchType.EAGER)
    @JoinColumn(name = "questionnaireId")
    val questions: Set<QuestionEntity>,
    @OneToMany(fetch = FetchType.EAGER)
    @JoinColumn(name = "questionnaireId")
    val answers: Set<AnswerEntity>
)

@Entity
@Table(name = "question")
data class QuestionEntity(
    @Id
    @GeneratedValue(generator = "UUID")
    @GenericGenerator(name = "UUID", strategy = "org.hibernate.id.UUIDGenerator")
    val id: UUID?,
    val optional: Boolean,
    val questionLabel: String?,
    val questionKey: String?,
    @Enumerated(EnumType.STRING)
    val questionType: QuestionType,
    @Enumerated(EnumType.STRING)
    val answerType: AnswerType,
    @OneToMany(fetch = FetchType.EAGER)
    @JoinColumn(name = "questionId")
    val answers: Set<AnswerEntity>,
    @OneToMany(fetch = FetchType.EAGER)
    @JoinColumn(name = "questionId")
    val branches: Set<BranchEntity>
)

@Entity
@Table(name = "answer")
@TypeDefs(
    TypeDef(name = "jsonb", typeClass = JsonBinaryType::class)
)
data class AnswerEntity(
    @Id
    @GeneratedValue(generator = "UUID")
    @GenericGenerator(name = "UUID", strategy = "org.hibernate.id.UUIDGenerator")
    val id: UUID?,
    val questionId: UUID,
    @Type(type = "jsonb")
    val value: Map<String, Any>?,
    val order: Int,
    @OneToMany(fetch = FetchType.EAGER)
    @JoinColumn(name = "answerId")
    val validations: Set<AnswerValidationEntity>
)

Answer 1

这些方法行为不同的原因是因为它们的实现非常不同。

FindByID是使用EntityManager.find实现的。这是可能的，因为它只在SimpleJParepository中实现了一次，它实现了CrudRepository的方法。EntityManager.find反过来支持实体上的惰性/热切配置。

当使用某种形式的查询时，该配置被忽略，这正是添加到存储库中的其他findby...方法所发生的情况。您可以用@entitygraph注释它们，以证明一个实体图，指定您希望它使用的获取策略。