Spring Data JPA 在保存时通过 id 进行许多选择

Question

我有这个模型：

数据模型：

@Table(name = "data_model")
public class DataModel {

    @Id
    @GeneratedValue
    @Column(name = "model_id")
    private Integer id;

    @OneToMany(orphanRemoval = true, mappedBy = "dataModel", cascade = {CascadeType.PERSIST, CascadeType.REMOVE, CascadeType.MERGE})
    private List<OutputField> outputFields;

    @OneToMany(orphanRemoval = true, mappedBy = "dataModel", cascade = {CascadeType.PERSIST, CascadeType.REMOVE, CascadeType.MERGE})
    private List<IncrementField> incrementFields;

}

输出字段：

@Table(name = "data_model_output_field")
public class OutputField {

    @Id
    @GeneratedValue
    @Column(name = "output_field_id")
    private Integer id;

    @ManyToOne(optional = false, fetch = FetchType.LAZY, cascade = CascadeType.REFRESH)
    @JoinColumn(name = "model_id")
    private DataModel dataModel;

}

增量字段：

@Entity
@Table(name = "data_model_increment_field")
public class IncrementField {

    @Id
    @GeneratedValue
    @Column(name = "increment_field_id")
    private Integer id;
    
    @ManyToOne(optional = false, fetch = FetchType.LAZY, cascade = CascadeType.REFRESH)
    @JoinColumn(name = "model_id")
    private DataModel dataModel;
}

我有服务方法：

@Transactional
public void createOrUpdate(DataModel dataModel) {
    DataModel savedDataModel = dataModelRepository.save(dataModel);
}

当我尝试保存实体时，在日志中我看到许多选择查询：

Name:test-ds, Connection:9, Time:0, Success:True
Type:Prepared, Batch:False, QuerySize:1, BatchSize:0
Query:["
    select
        outputfiel0_.model_id as model_id8_4_0_
    from
        data_model_output_field outputfiel0_ 
    where
        outputfiel0_.output_field_id=1"]


Name:test-ds, Connection:9, Time:0, Success:True
Type:Prepared, Batch:False, QuerySize:1, BatchSize:0
Query:["
    select
        outputfiel0_.model_id as model_id8_4_0_
    from
        data_model_output_field outputfiel0_ 
    where
        outputfiel0_.output_field_id=2"]


Name:test-ds, Connection:9, Time:0, Success:True
Type:Prepared, Batch:False, QuerySize:1, BatchSize:0
Query:["
    select
        outputfiel0_.model_id as model_id8_4_0_
    from
        data_model_output_field outputfiel0_ 
    where
        outputfiel0_.output_field_id=3"]


Name:test-ds, Connection:9, Time:0, Success:True
Type:Prepared, Batch:False, QuerySize:1, BatchSize:0
Query:["
    select
        outputfiel0_.model_id as model_id8_4_0_
    from
        data_model_output_field outputfiel0_ 
    where
        outputfiel0_.output_field_id=4"]

Name:test-ds, Connection:9, Time:0, Success:True
Type:Prepared, Batch:False, QuerySize:1, BatchSize:0
Query:["
    select
        outputfiel0_.model_id as model_id8_4_0_
    from
        data_model_output_field outputfiel0_ 
    where
        outputfiel0_.model_id=2"]

它通过 id 获取每个字段。但是他为什么这样做，他可以立即通过model_id拿起名单？

此外，在日志的最后，他做到了。那他为什么需要之前的查询呢？

更新：

重现的简单示例：https://github.com/eaxdev/simple-web-app

Answer 1

那是因为save 方法使用EntityManager.merge 刷新更改，而您正在使用 MERGE 级联。我不知道您要在此处保存哪种对象图，但似乎 outputFields 集合有 4 个条目首先经历 MERGE 级联，即 Hibernate 选择这些对象的当前状态，以便它可以刷新更改。接下来，Hibernate 尝试刷新 DataModel#outputFields 关系的更改，即外键，它需要查询所有对象，以便知道要删除什么（我认为这是由于孤立删除而需要的）。

在我看来OutputField 和IncrementField 与DataModel 有组合关系，那么为什么不将它们建模为@Embeddable？这样，Hibernate 可以更有效地管理关联。