Question

我有一个有多对多关系的实体：
有Product，可以有许多类别和许多标签。 Tag或Category也可以有很多产品。

现在的架构是：

products
-------------
product_id
name

categories
-------------
category_id
name

tags
-------------
tag_id
name

products_categories
-------------
product_id
category_id

products_tags
-------------
product_id
tag_id

我需要在一个页面中显示包含所有标签和类别的产品列表问题是我执行的查询数量是2n + 1，其中n是页面中列出的产品数量。

基本上我先做的是：

select product_id, name from products

然后在我的应用程序的循环中，我执行这些查询（伪代码）：

tags = []
categories = []
for each product
    tags[product_id] = select t.tag_id, t.name 
                       from products_tags as pt 
                       join tags t 
                           on pt.tag_id = t.tag_id
                       where pt.product_id = {product_id}

    categories[product_id] = select c.category_id, c.name 
                             from products_categories as pc 
                             join categories c 
                                 on pc.category = c.category_id
                             where pc.product_id = {product_id}
end for each

是否有一种很好的方法可以使执行的查询数量与查询的记录数无关？

的修改
对于每种产品，我需要以这种格式显示数据：

-------------------------------------------------------
| Product name: A good smartphone                     |
| Categories: Tech, Smartphone                        |
| Tags: smartphone, some-smartphone-brand, 4g         |
|                                                     |
-------------------------------------------------------

Answer 1

你可以试试这个（它依次使用带有标签和类别的连接，并用列table_type来区分它们）：

select *, 'CAT' as table_type
from products p left join products_categories pc left join categories c  
ON p.product_id = pc.product_id and pc.category_id = c.category_id
UNION ALL
select *, 'TAG' as table_type
from products p left join products_tags pt left join tags t  
ON p.product_id = pt.product_id and pt.tag_id = t.tag_id
order by p.product_id

Answer 2

您可以使用单个查询从数据库和应用程序级别获取所有信息，根据您的需要构建数据。下面的查询将多次返回每个产品，但标签/类别不同。

select p.product_id, p.name, t.tag_id, t.name tag_name, c.category_id, c.cat_name 
from product p
join products_tags as pt on pt.product_id = p.product_id
join tags t on pt.tag_id = t.tag_id
join products_categories as pc on pc.product_id = p.product_id
join categories c on pc.category = c.category_id
order by p.product_id

伪代码

products = []
tags = []
categories = []
var productObj= {};
var product_id =false;
for each product
    /**
    * If product_id != product.id from loop iteration it means this is different product from previous iteration, Push this product object in `products[]`
    */
    if(product.id != product_id){
        products[product.id] = product;
        product_id = product.id;
    }
    /** you can put additional check here before insert in array if value is not present  `products[product.id][tags]` */
    products[product.id][tags][] = tag_name;

    /** you can put additional check here before insert in array if value is not present  `products[product.id][categories]` */ 
    products[product.id][categories][]= cat_name;
end for each

产品的样本对象/数据结构

[ 
    1 => [
    name => A good smartphone,
    categories => [Tech, Smartphone],
    tags => [smartphone, some-smartphone-brand, 4g]
    ],  
    2 => [
    name: Another product,
    categories => [Cat1, Cat2],
    tags => [Tag1, Tag2, .....]
    ],
.....
]

使用多对多关系使查询量保持不变

2 个答案: