This document discusses Spring JPA and solutions to the N+1 problem when querying related entities. It introduces Spring JPA and concepts like the EntityManager and entity relationships. The N+1 problem occurs when querying a parent entity, like Cluster, which then requires separate queries for each related child entity, like Database. Solutions include lazy loading child entities only when needed, using joins to retrieve related data in a single query, and entity graphs to customize entity fetching.

1. Spring JPA
技术分享
2019-09

2. Entity及其关系2
N+1问题及其解决方案3
JPA简介1

3. JPA简介1

4. 基本概念
• JPA（Java Persistence API）
• Hibernate
• Spring Data JPA

5. Entity的生命周期

6. EntityManager
Java Object Record
EntityManager
Create
Retrieve
Update
Delete
persist()
find()
merge(),flush()
remove()

7. Entity及其关系2

8. Entity及其关系
关系：
• 一对一
• 一对多
• 多对多
关系类型：
• 单向
• 双向

9. 代码表示
注释类型：
• @OneToOne
• @ManyToOne
• @OneToMany
• @ManyToMany
@Entity
@Table(name = "`database`")
public class Database {
private Long id;
private Cluster cluster;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "`id`")
public Long getId() {
return id;
}
@ManyToOne(fetch = FetchType.LAZY, optional = false)
@JoinColumn(name = "`cluster_id`", nullable = false)
public Cluster getCluster() {
return cluster;
}
}

10. 注释属性的作用
@Target({METHOD, FIELD})
@Retention(RUNTIME)
public @interface OneToMany {
Class targetEntity() default void.class;
CascadeType[] cascade() default {};
FetchType fetch() default LAZY;
String mappedBy() default "";
boolean orphanRemoval() default false;
}

11. 3 N+1问题及其解决方案

12. N+1问题
在查询Cluster时，查到n条结果，
每个结果还会附加一次Database
查询。
@Entity
@Table(name = "`cluster`")
public class Cluster {
private Long id;
private List<Database> databases;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "`id`")
public Long getId() {
return id;
}
@OneToMany(fetch = FetchType.EAGER,
mappedBy = "`database_id`")
public List<Database> getDatabase() {
return databases;
}
}
select * from cluster;
+
select * from database where cluster_id = ?
select * from database where cluster_id = ?
…

13. 解决方案
1. 使用懒加载，用到
database的时候再去
查询。
@Entity
@Table(name = "`cluster`")
public class Cluster {
private Long id;
private List<Database> databases;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "`id`")
public Long getId() {
return id;
}
@OneToMany(fetch = FetchType.LAZY , mappedBy = "`database_id`")
public List<Database> getDatabase() {
return databases;
}
}

14. 解决方案
2. 使用Join的方式
public interface ClusterRepository extends CrudRepository<Cluster,Long> {
@Override
@Query(value = "select * from cluster left outer join database
on id = database.cluster_id", nativeQuery = true)
Iterable<Cluster> findAll();
}

15. 解决方案
3. 使用Join的方式(EntityGraph)
@Entity
@Table(name = "`cluster`")
@NamedEntityGraph(name = “cluster-entity-graph",
attributeNodes = {@NamedAttributeNode(“databases")})
public class Cluster {
private Long id;
private List<Database> databases;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "`id`")
public Long getId() {
return id;
}
@OneToMany(fetch = FetchType.LAZY , mappedBy
="`database_id`")
public List<Database> getDatabase() {
return databases;
}
}
select * from cluster left outer join database
on id = database.cluster_id
public interface ClusterRepository extends
CrudRepository<Cluster, Integer> {
@Override
@EntityGraph("cluster-entity-graph")
Iterable<Cluster> findAll();
}

16. EntityGraph的好处
可以在运行时调整取用实体的方式，更加灵活。
EntityGraph entityGraph = entityManager.getEntityGraph("user-entity-graph");
Map<String, Object> properties = new HashMap<>();
properties.put("javax.persistence.fetchgraph", entityGraph);
entityManager.find(User.class, 1,properties);
entityManager.find(User.class, 1,properties); Lazy方式
Join方式