JPQL 如何总结给定的属性个子项并获取这些子项避免 n + 1 个选择？

Question

我有两个实体：类别和产品。它们是关联的并且类别是父级：

@Entity
@Table(name = "categories")
public class Category {
    @Id
    @GeneratedValue(generator = "inc")
    @GenericGenerator(name = "inc", strategy = "increment")
    private int id;
    private String name;
    private int totalQuantity;

    @OneToMany(fetch = FetchType.LAZY, mappedBy = "category")
    private Set<Product> products;
  
    public Category(int id, String name, int totalQuantity, Set<Product> products) {
        this.id = id;
        this.name = name;
        this.totalQuantity = totalQuantity;
        this.products = products;
    }

产品实体：

@Entity
@Table(name = "products")
public class Product {
    @Id
    @GeneratedValue(generator = "inc")
    @GenericGenerator(name = "inc", strategy = "increment")
    private int id;
    private String name;
    private int amount;
    @ManyToOne
    @JoinColumn(name = "category_id")
    private Category category
}

（totalQuantity 是与类别关联的 amount 产品的总和）我想以防止 n + 1 并进行求和的方式获取所有类别和所有相关产品。这是我的查询 wrong/uncompleted 因为我不知道如何 do/complete 它：

@Query("SELECT new com.example.demo.category.Category(p.category.id, p.category.name, SUM(p.amount), ) FROM Product p GROUP BY p.category.id")
List<Category> findAll();

编辑：为了更好地展示目标，我添加了类别视图（“前端”、“后端”）以及与每个类别相关联的产品：

Answer 1

我有一个项目有这个功能，你可以像这样创建一个实体

SampleEntity

@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "sample")
    private Set<AttributeEntity> attributes = new HashSet<>();

带有一个实体列表。创建一个包含长期字段的 DTO，如

SampleDTO

当您将实体转换为 DTO 对象时，您可以使用计数，在我的例子中，我使用了 mapStruct 插件

SampleMapper


    public abstract List<SampleDTO> convert(List<SampleEntity> sampleEntities);

    public long map(Set<AttributeEntity> past) {
        return past.size();
    }

在我的服务中，我调用 find all

@Override
    public List<SampleDTO> findAll() {
        List<SampleEntity> entities = sampleRepository.findAll();
        return sampleMapper.convert(entities);
    }

并且存储库是 jpaRepository

的默认存储库


public interface ISampleRepository extends JpaRepository<SampleEntity, UUID> , JpaSpecificationExecutor<SampleEntity> {

}

Answer 2

使用@Formula关键字做原生sql求和产品金额。

// the p.category_id=id <-- this id is Category itself id
@Formula("(SELECT COALESCE(SUM(p.amount),0) FROM products p INNER JOIN categories c ON p.category_id=c.id WHERE p.category_id=id)")
private int totalQuantity;

Note: If your totalQuantity type is int, you need to use COALESCE avoid the null value. If it type is Integer, you don't need to use it.

使用LEFT JOIN FETCH防止N+1问题。

@Query("SELECT DISTINCT c FROM Category c LEFT JOIN FETCH c.products")
List<Category> findAll()

Hibernate sql 结果：

select distinct category0_.id                         as id1_6_0_,
                products1_.id                         as id1_11_1_,
                category0_.name                       as name2_6_0_,
                (SELECT COALESCE(SUM(p.amount), 0)
                 FROM product p
                          INNER JOIN categories c ON p.category_id = c.id
                 where p.category_id = category0_.id) as formula1_0_,
                products1_.amount                     as amount2_11_1_,
                products1_.category_id                as category4_11_1_,
                products1_.name                       as name3_11_1_,
                products1_.category_id                as category4_11_0__,
                products1_.id                         as id1_11_0__
from categories category0_
         left outer join products products1_ on category0_.id = products1_.category_id

您可以使用一个查询同时获取所有类别和产品总金额。

旧答案

您可以使用 LEFT JOIN FETCH 来防止 n+1 问题。

@Query("SELECT DISTINCT c FROM Category c LEFT JOIN FETCH c.products")
List<Category> findAll();

使用 sum() 运算符计算产品数量。

List<Category> categories = categoryRepository.findAll().stream().peek(category -> {
    Set<Product> products = category.getProducts();
    if (!ObjectUtils.isEmpty(products)) {
        int totalQuantity = products.stream().mapToInt(Product::getAmount).sum();
        category.setTotalQuantity(totalQuantity);
    }
}).collect(Collectors.toList());

最后，您可以使用一个查询来获取所有类别并统计产品的总量。

JPQL 如何总结给定的属性个子项并获取这些子项避免 n + 1 个选择？

JPQL How to sum up a given property of children and fetch these children avoiding n + 1 selections?

jpa

jpql

spring-boot

旧答案

JPQL 如何总结给定的 属性 个子项并获取这些子项避免 n + 1 个选择？

JPQL How to sum up a given property of children and fetch these children avoiding n + 1 selections?

jpa

jpql

spring-boot

旧答案

JPQL 如何总结给定的属性个子项并获取这些子项避免 n + 1 个选择？