1つの結果でさらにテーブルに結合する

Question

私は、乗算テーブルからsql selectのヘルプが必要です。

テーブル：

s_products 
- id 
- name 
- url 
- brand_id 

s_products_categories 
- product_id 
- category_id 

s_brands 
- id 
- name 
- url 

s_categories 
- id 
- name 
- url 

が比較：

s_products.id = s_products_categories.product_id 
s_categories.id = s_products_categories.category_id 
s_products.brand_id = s_brands.id 

し、必要な結果：

1）$products['s_categories.url']['s_brands.url']['s_products.url'] = 's_products.name';

2）$brands['s_categories.url']['s_brands.url'] = 's_brands.name';

あなたは1つまたは複数の選択肢でそれをやることを手伝ってもらえますか？

Answer 1

試してみてください（。例えばbrand_urlでcategory_url）

select c.url category_url, 
     c.name category_name, 
     b.url brand_url, 
     b.name brand_name, 
     p.url product_url, 
     p.name product_name 
from s_products p 
join s_brands b on p.brand_id = b.id 
join s_products_categories pc on p.id = pc.product_id 
join s_categories c on pc.category_id = c.id 

特定のフィールドの組み合わせを表示するには、すぐにselect後distinctを追加し、要求されないフィールドを削除。

Answer 2

select *yourcolumnsname* 
from s_products INNER JOIN s_products_categories 
on s_products.id = s_products_categories.product_id 
INNER JOIN s_categories.id 
on s_categories.id = s_products_categories.category_id; 

Answer 3

SELECT s_categories.url as url1, s_products.url as url2, s_products.name as name 
FROM s_products 
INNER JOIN s_products_categories on s_products.id = s_products_categories.product_id 
INNER JOIN s_categories   on s_categories.id = s_products_categories.category_id