Sql 如何查找表中在给定列表中唯一具有相关行的所有行？

我有三张桌子。可以这样想：

  Recipes
  id | name
  1  | Cookies
  2  | Soup
  ...


  Ingredients
  id | name
  1  | flour
  2  | butter
  3  | chicken
  ...

  Recipe_Ingredient
  recipe_id | ingredient_id
  1         | 1
  1         | 2
  2         | 3

希望你能明白。我想要的是一个查询，在那里我可以找到所有配方，其中的成分是一组给定成分的子集

我的想法是，我希望有一份清单，列出所有我手头上的东西（当然不是所有我手头上的东西）

---

我尝试用不同级别的子查询和与EXISTS相关的子查询来实现这一点，但没有成功。我也尝试过使用HAVING和COUNT，但这似乎只适用于我想要的东西使用了我手头上所有的成分。

如果有效，测试一下。假设您的（唯一）配料id在名为“配料”的表中可用，此查询应显示包含完整配料的配方id：

    mysql>
    mysql> select * from ingredients;
    +------+---------------+-----------+
    | id   | name          | available |
    +------+---------------+-----------+
    |    1 | salt          | n         |
    |    2 | sugar         | n         |
    |    3 | flour         | n         |
    |    4 | butter        | n         |
    |    5 | vanilla       | n         |
    |    6 | baking powder | n         |
    |    7 | egg           | n         |
    +------+---------------+-----------+
    7 rows in set (0.00 sec)

    mysql> select * from recipes;
    +------+---------------+
    | id   | name          |
    +------+---------------+
    |    1 | cookie        |
    |    2 | soup          |
    |    3 | xtreme flavor |
    +------+---------------+
    3 rows in set (0.00 sec)

    mysql> select * from recipe_ingredient;
    +-----------+---------------+
    | recipe_id | ingredient_id |
    +-----------+---------------+
    |         1 |             1 |
    |         1 |             2 |
    |         1 |             3 |
    |         1 |             4 |
    |         1 |             5 |
    |         1 |             6 |
    |         1 |             7 |
    |         2 |             1 |
    |         2 |             7 |
    |         3 |             4 |
    |         3 |             3 |
    +-----------+---------------+
    11 rows in set (0.00 sec)

    mysql>
    mysql> update ingredients set available = 'n';
    Query OK, 0 rows affected (0.00 sec)
    Rows matched: 7  Changed: 0  Warnings: 0

    mysql>
    mysql> update ingredients set available = 'y'
        -> where id in  (1,2,3,4,5,6,7);
    Query OK, 7 rows affected (0.00 sec)
    Rows matched: 7  Changed: 7  Warnings: 0

    mysql>
    mysql> select recipes.name from
        -> (select recipe_id, available from
        -> recipe_ingredient,
        -> ingredients
        -> where ingredient_id = ingredients.id
        -> group by  recipe_id, available) x, recipes
        -> where recipes.id = x.recipe_id
        -> group by x.recipe_id
        -> having count(*) = 1
        -> and max(x.available) = 'y';
    +---------------+
    | name          |
    +---------------+
    | cookie        |
    | soup          |
    | xtreme flavor |
    +---------------+
    3 rows in set (0.06 sec)

    mysql>
    mysql> update ingredients set available = 'n';
    Query OK, 7 rows affected (0.00 sec)
    Rows matched: 7  Changed: 7  Warnings: 0

    mysql>
    mysql> update ingredients set available = 'y'
        -> where id in  (1,7);
    Query OK, 2 rows affected (0.00 sec)
    Rows matched: 2  Changed: 2  Warnings: 0

    mysql>
    mysql> select recipes.name from
        -> (select recipe_id, available from
        -> recipe_ingredient,
        -> ingredients
        -> where ingredient_id = ingredients.id
        -> group by  recipe_id, available) x, recipes
        -> where recipes.id = x.recipe_id
        -> group by x.recipe_id
        -> having count(*) = 1
        -> and max(x.available) = 'y';
    +------+
    | name |
    +------+
    | soup |
    +------+
    1 row in set (0.06 sec)

    mysql>
    mysql>
    mysql> update ingredients set available = 'n';
    Query OK, 2 rows affected (0.00 sec)
    Rows matched: 7  Changed: 2  Warnings: 0

    mysql>
    mysql> update ingredients set available = 'y'
        -> where id in  (4,3);
    Query OK, 2 rows affected (0.00 sec)
    Rows matched: 2  Changed: 2  Warnings: 0

    mysql>
    mysql> select recipes.name from
        -> (select recipe_id, available from
        -> recipe_ingredient,
        -> ingredients
        -> where ingredient_id = ingredients.id
        -> group by  recipe_id, available) x, recipes
        -> where recipes.id = x.recipe_id
        -> group by x.recipe_id
        -> having count(*) = 1
        -> and max(x.available) = 'y';
    +---------------+
    | name          |
    +---------------+
    | xtreme flavor |
    +---------------+
    1 row in set (0.05 sec)

    mysql>
    mysql>
    mysql> update ingredients set available = 'n';
    Query OK, 3 rows affected (0.00 sec)
    Rows matched: 7  Changed: 3  Warnings: 0

    mysql>
    mysql> update ingredients set available = 'y'
        -> where id in  (1,3,7);
    Query OK, 3 rows affected (0.00 sec)
    Rows matched: 3  Changed: 3  Warnings: 0

    mysql>
    mysql> select recipes.name from
        -> (select recipe_id, available from
        -> recipe_ingredient,
        -> ingredients
        -> where ingredient_id = ingredients.id
        -> group by  recipe_id, available) x, recipes
        -> where recipes.id = x.recipe_id
        -> group by x.recipe_id
        -> having count(*) = 1
        -> and max(x.available) = 'y';
    +------+
    | name |
    +------+
    | soup |
    +------+
    1 row in set (0.06 sec)

    SELECT recipe_id
    FROM [select recipe_id, count(*) as num_of_ingredients from recipe_ingredient group by recipe_id]. AS x, [select recipe_ingredient.recipe_id as recipe_id, count(*) as num_of_ingredients 
    from recipe_ingredient, ingredients_avail               
    where               
     recipe_ingredient.ingredient_id = ingredients_avail.ingredient_id
    group by recipe_ingredient.recipe_id]. AS y               
    WHERE x.recipe_id = y.recipe_id and               
    x.num_of_ingredients = y.num_of_ingredients;               

此外，这里的语法更为典型：

  SELECT x.recipe_id 
    FROM ( 
      SELECT recipe_id, count(*) as num_of_ingredients 
        FROM recipe_ingredients
      GROUP BY recipe_id
     ) x, ( 
      SELECT recipe_id, count(*) as num_of_ingredients 
        FROM recipe_ingredients
        JOIN ingredients_avail 
          ON recipe_ingredients.ingredient_id = ingredients_avail.ingredient_id 
      GROUP BY recipe_id
     ) y 
  WHERE x.recipe_id = y.recipe_id AND x.num_of_ingredients = y.num_of_ingredients;

---

因此，你希望食谱中的所有成分都在“可用”成分列表中。试试[sql match all]标签或右边相关标题下的链接。这叫做@ypercube哇，太棒了。是的，我就是这么想的，但我一定是把我的逻辑搞砸了。谢谢你的精彩链接。这个问题有10多种方法可以达到这个目的。也进行了基准测试。更一般的答案是1,2,7,8（如果您的可用成分在表或列表中，并且您不想动态创建SQL查询）。对，最终这些将在表中。如果我有可用的成分1,3和7，这是否正确工作？也就是说，它应该返回汤，因为我可以用1和7来制作汤。@Frew，是的，当1、3和7可用时，它会返回“汤”。我在上面添加了结果。太棒了，谢谢；我会看看这对我是否有效，如果是的话，你会得到奖励：-）事实上，我在想如何将其转换为我的模式时遇到了很多困难；也就是说，ID列表或子查询，而不是可用内容的标志列。我能得到一些帮助吗？