Mysql SQL：给定特定元素的精确行匹配

这是我第一次在这里发帖。我似乎找不到我问题的答案

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，它只展示了可以用“书架”上现有元素制作的食谱。 这些成分必须有精确的成分匹配

用户：

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 |
+---------+------------+---------------+

这是我的配方和配料关系表

配方和配料：

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

我尝试了以下查询：

    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'

将返回此表：

+--------+------------+--------------+
| 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 |
+--------+------------+--------------+

虽然配方1确实包含我书架上的配料，但它也缺少配料10

我只想得到这种结果集

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

因为配方2包含了我书架上的所有元素

---

我整天都在处理这个问题。。。可能会缺少什么？

这有点棘手，因为您需要按配方聚合当前查询，但还需要原始查询以获取完整记录。遗憾的是，MySQL不支持公共表表达式或其他特性，这可能会减少查询的详细程度

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'

这里的基本策略是只对一个子查询执行一个附加连接，该子查询标识每个成分都属于给定用户的所有配方。关键部分如下：

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

---

这将检查给定配方的总行数是否与分配给给定用户的行数相匹配。

请尝试发布

usr

和

shf

表，以获得更清晰的信息。我知道这个项目不会像看上去那么简单！谢谢让我更仔细地分析和理解这个答案。这个方法是否需要应用于配方而不是用户？@Kerberos

COUNT（usr.email）

统计给定用户的非

NULL

（即匹配）成分的数量。

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'

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