解决 SpringData JPA 的n+1问题

1. 首先解决 n+1 问题

（1）Entity 添加 @NamedEntityGraph 

@Entity
@Table(name = "tb_depart_detail", schema = "")
@NamedEntityGraph(name = "depart_detail.Graph", attributeNodes = {
        @NamedAttributeNode(value = "depart")
})
@SuppressWarnings("serial")
public class DepartDetailEntity implements java.io.Serializable {
    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "depart_id")
    public DepartEntity getDepart() {
        return depart;
    }

    public void setDepart(DepartEntity depart) {
        this.depart = depart;
    }
}

（2） 重写 JpaRepository 的API 指定使用 NameEntityGraph

import com.doctor.assistant.userserver.springdata.entity.DepartDetailEntity;
import org.springframework.data.jpa.repository.EntityGraph;
import org.springframework.data.jpa.repository.JpaRepository;

import java.util.List;

public interface DepartDetailRepository extends JpaRepository<DepartDetailEntity, String> {

    @Override
    @EntityGraph(value = "depart_detail.Graph",type = EntityGraph.EntityGraphType.FETCH)
    List<DepartDetailEntity> findAll();
}

（3） Test 测试 ：

2. 三层及更多关联时，解决 n+1 的方式：

@Entity
@Table(name = "t_base")
@Inheritance(strategy = InheritanceType.JOINED)
@NamedEntityGraph(name = "base.Graph", attributeNodes = {
        @NamedAttributeNode(value = "someList"),
        @NamedAttributeNode(value = "userList", subgraph = "outterGraph")
}
,subgraphs = {
        @NamedSubgraph(name = "outterGraph",attributeNodes = {
                @NamedAttributeNode(value = "outterList",subgraph = "innerGraph")
        }),@NamedSubgraph(name = "innerGraph",attributeNodes = {
                @NamedAttributeNode("innerList")
        })
})

3. 若想保留原接口，再写一个 Repository的实现类，其中什么也不做，即可保留原方法：

@Repository
public interface EAGERUserRepository extends JpaRepository<TSBaseUser, String> {
}