我有三张桌子,
Product_to_categories => contains two columns (Category_id, Product_id)
Product => contains a few columns (product_id, name, sku, ...)
categories => contains a few columns (category_id, name, ....)
我想得到一个结果集,前10个独特的category_id与产品表一起加入,一个类别可以分配许多产品,我想要退回一个类别中的所有产品,但我只想得到前8个类别的产品......
当前查询:
SELECT p2c.category_id, p.pname, c.category_name
FROM product p LEFT JOIN product_to_category p2c
ON (p.product_id = p2c.product_id)
Left Join category c (p2c.category_id = c.category) LIMIT 0,8
当前输出
catID | p.name | catName |<br/>
1 | docs | shoe<br/>
1 | bob | shoe<br/>
1 | mom | shoe<br/>
1 | cat | shoe<br/>
1 | dang | shoe<br/>
1 | kit | shoe<br/>
2 | pis | book<br/>
2 | jiz | book<br/>
目前我只得到前8个结果无意义,但我希望获得以下输出:
catID | p.name | catName |<br/>
1 | docs | shoe<br/>
1 | bob | shoe<br/>
1 | mom | shoe<br/>
1 | cat | shoe<br/>
1 | dang | shoe<br/>
1 | kit | shoe<br/>
2 | pis | book<br/>
2 | jiz | book<br/>
3 | docs | shirt<br/>
3 | bob | shirt<br/>
3 | mom | shirt<br/>
4 | cat | light<br/>
4 | dang | light<br/>
5 | kit | sound<br/>
6 | pis | mic<br/>
6 | jiz | mic<br/>
7 | docs | pen<br/>
7 | bob | pen<br/>
7 | mom | pen<br/>
7 | cat | pen<br/>
8 | dang | lace<br/>
8 | kit | lace<br/>
8 | pis | lace<br/>
8 | jiz | lace<br/>
我希望结果集包含分配给前8个类别的所有产品的结果...
请指教。 谢谢 哈迪
答案 0 :(得分:0)
我有点不确定为什么你的查询是从产品表开始的,所以我不打算 - 但除此之外,我认为你需要的是:
SELECT c.category_id, p.pname, c.category_name
FROM category c
INNER JOIN product_to_category pc ON c.category_id = pc.category_id
INNER JOIN product p ON pc.product_id = p.product_id
WHERE c.category_id in (SELECT TOP 8 category_id FROM Category)
ORDER BY 1, 2