Spring 數據 JPA 保存時通過 id 進行多次選擇

Question

我有這個模型：

數據模型：

@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;

}

輸出字段：

@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;
}

我有服務方法：

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

當我嘗試保存實體時，在日志中我看到許多 select 查詢：

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"]

它通過 id 獲取每個字段。 但是他為什么這樣做，他可以立即通過model_id拿起列表？

此外，在日志的最后，他做到了。 為什么他需要以前的查詢呢？

更新：

復制的簡單示例： https://github.com/eaxdev/simple-web-app

Answer 1

那是因為save方法使用EntityManager.merge刷新更改，而您正在使用 MERGE 級聯。 我不知道您要在此處保存什么樣的 object 圖，但似乎outputFields集合有 4 個條目首先經歷 MERGE 級聯，即 Hibernate 選擇當前 Z9ED39E2EA9314586B6EZA985 的這些對象。變化。 接下來，Hibernate 嘗試刷新DataModel#outputFields關系的更改，即外鍵，它需要查詢所有對象，以便它知道要刪除什么（我認為這是由於孤立刪除而需要的）。

在我看來， OutputField和IncrementField與DataModel有組合關系，那么為什么不把 model 當作@Embeddable呢？ 這樣，Hibernate 可以更有效地管理關聯。