mysql - SQL:给定某些元素的精确行匹配

Question

这是我第一次在这里发帖。我似乎没有找到问题的答案。

所以...我正在为一个学校项目安排一个数据库，这是一本食谱，仅显示可以使用“架子”中的现有元素制作的食谱。 这些成分必须具有精确的成分匹配。

用户:

+---------------+------+----------+----------+
|     email     | name | lastname | password |
+---------------+------+----------+----------+
| pal@mail.com  | John | Potato   | password |
| they@mail.com | Mary | Carrot   | password |
+---------------+------+----------+----------+

货架:

+---------+------------+---------------+
| shelfID | ingredient |     user      |
+---------+------------+---------------+
|       1 |          1 | pal@mail.com  |
|       2 |          2 | pal@mail.com  |
|       3 |          3 | pal@mail.com  |
|       4 |          4 | pal@mail.com  |
|       5 |         10 | they@mail.com |
|       6 |         12 | they@mail.com |
+---------+------------+---------------+

这是我的recipe_ingredient关系表

食谱成分:

+--------+------------+
| recipe | ingredient |
+--------+------------+
|      1 |          1 |
|      1 |          2 |
|      1 |          4 |
|      1 |         10 |
|      2 |          1 |
|      2 |          2 |
|      2 |          3 |
|      2 |          4 |
|      3 |          2 |
|      3 |          3 |
|      3 |         15 |
+--------+------------+

我尝试过这个查询:

    SELECT 
    rec_ing.recipe, shf.ingredient, shf.user
FROM
    recipes_ingredients AS rec_ing
        INNER JOIN
    shelf AS shf ON rec_ing.ingredient = shf.ingredient
        INNER JOIN
    users AS usr ON shf.user = usr.email
WHERE
    usr.email = 'pal@mail.com'

返回此表:

+--------+------------+--------------+
| recipe | ingredient |     user     |
+--------+------------+--------------+
|      1 |          1 | pal@mail.com |
|      1 |          2 | pal@mail.com |
|      1 |          4 | pal@mail.com |
|      2 |          1 | pal@mail.com |
|      2 |          2 | pal@mail.com |
|      2 |          3 | pal@mail.com |
|      2 |          4 | pal@mail.com |
|      3 |          2 | pal@mail.com |
|      3 |          3 | pal@mail.com |
+--------+------------+--------------+

虽然食谱 1 确实包含我架子上的成分，但它也缺少成分 10

+--------+------------+
| recipe | ingredient |
+--------+------------+
|      1 |          1 |
|      1 |          2 |
|      1 |          4 |
|      1 |         10 |
|    ... |        ... |
+--------+------------+

我试图只获得这种结果集。

+--------+------------+
| recipe | ingredient |
+--------+------------+
|      2 |          1 |
|      2 |          2 |
|      2 |          3 |
|      2 |          4 |
+--------+------------+

因为食谱 2 包含我的书架中的所有元素

我整天都在解决这个问题......可能缺少什么？

Answer 1

这有点棘手，因为您需要按配方聚合当前查询，但您还需要原始查询来获取完整记录。遗憾的是，MySQL 不支持公共(public)表表达式或其他可以为我们提供更简洁查询的功能。

SELECT 
    rec_ing.recipe,
    shf.ingredient,
    shf.user
FROM recipes_ingredients AS rec_ing
INNER JOIN shelf AS shf
    ON rec_ing.ingredient = shf.ingredient
INNER JOIN users AS usr
    ON shf.user = usr.email
INNER JOIN
(
    SELECT rec_ing.recipe
    FROM recipes_ingredients AS rec_ing
    INNER JOIN shelf AS shf
        ON rec_ing.ingredient = shf.ingredient
    LEFT JOIN users AS usr
        ON shf.user = usr.email
    WHERE
        usr.email = 'pal@mail.com'
    GROUP BY rec_ing.recipe
    HAVING COUNT(usr.email) = COUNT(*)
) t
    ON rec_ing.recipe = t.recipe
WHERE
    usr.email = 'pal@mail.com'

这里的基本策略是仅对子查询进行一次附加联接，该子查询标识其中每种成分属于给定用户的所有食谱。关键部分如下:

HAVING COUNT(usr.email) = COUNT(*)

这会检查给定配方的总行数是否与已分配给给定用户的行数匹配。