SQL 查询 INNER JOIN 仅当全部存在时

Question

我有两张表，一张包含不同的食谱，另一张包含 recipes_ingredients（食谱所需的所有成分）。

---

recipes
-----
id_recipe
name
description
etc...

recipe_ingredients
-----
id_recipe_ingredient
id_recipe
id_ingredient

---

假设我只选择了一种成分（例如鸡蛋）。

SELECT * 
FROM recipes r 
  INNER JOIN recipes_ingredients ri ON r.id_recipe = ri.id_recipe 
WHERE ri.id_ingredient = 1 

这将返回所有包含鸡蛋的食谱。这是合乎逻辑的，但我想创建一个查询，返回所有 ONLY 包含鸡蛋（例如煮鸡蛋）的食谱。不是一些包含鸡蛋的食谱。

---

如果我同时选择鸡蛋和牛奶并创建相同的查询：

SELECT * 
FROM recipes r 
  INNER JOIN recipes_ingredients ON r.id_recipe = ri.id_recipe 
WHERE ri.id_ingredient = 1 
  OR ri.id_ingredient = 2

我会找到煎蛋卷，但我也会找到所有其他包含鸡蛋和牛奶的食谱。

---

我要选择的是：

• 只包含鸡蛋的食谱

• 只包含鸡蛋和牛奶的食谱

我不想选择包含鸡蛋、牛奶、面粉和糖的蛋糕。

---

感谢您提供任何可能使我朝着正确方向前进的答案！

Answer 1

计算成分的数量和匹配成分的数量，并检查它们是否匹配。例如，要检索同时包含成分 1 和 2 且仅包含这两种成分的所有食谱：

SELECT id_recipe
FROM recipe_ingredients
GROUP BY id_recipe
HAVING COUNT(*) = 2
  AND COUNT(CASE WHEN id_ingredient IN (1, 2) THEN 1 END) = COUNT(*);

您还可以使用GROUP_CONCAT 将成分 ID 收集到一个字符串中并进行字符串比较：

SELECT id_recipe
FROM recipe_ingredients
GROUP BY id_recipe
HAVING GROUP_CONCAT(id_ingredient ORDER BY id_ingredient) = '1,2';

在前面的查询中，为了便于阅读，我省略了对配方的完整检索。为了完整起见，以下查询将检索完整的配方：

SELECT *
FROM recipes JOIN recipe_ingredients USING (id_recipe)
WHERE id_recipe in (
    SELECT id_recipe
    FROM recipe_ingredients
    GROUP BY id_recipe
    HAVING COUNT(*) = 2
      AND COUNT(CASE WHEN id_ingredient IN (1, 2) THEN 1 END) = COUNT(*)
);

虽然这些查询简短易读，但如果您有大量食谱和/或配料，则可能会遇到性能问题。如果这样做，您可能希望在 id_ingredient 和 id_recipe 上添加索引并尝试将查询重写为

SELECT *
FROM recipes JOIN recipe_ingredients USING (id_recipe)
WHERE id_recipe in (
    SELECT i1.id_recipe
    FROM recipe_ingredients i1
      JOIN recipe_ingredients i2
        ON (i1.id_recipe = i2.id_recipe AND i2.id_ingredient = 2)
      LEFT JOIN recipe_ingredients ni
        ON (i1.id_recipe = ni.id_recipe and ni.id_ingredient NOT IN (1, 2))
    WHERE i1.id_ingredient = 1 AND ni.id_ingredient IS NULL
);

但是，我发现此版本更麻烦且更难维护，因此建议在需要性能之前避免使用它。

Answer 2

用 EGG 和 MILK 的那些只有两种成分，UNION 到那些只有 EGG 和一种成分的........

SELECT r.Id, COUNT(ri.Id) FROM recipes r                            --GET THE RECIPE ID AND COUNT OF INGREDIENTS
INNER JOIN recipes_ingredients ri ON r.id_recipe = ri.id_recipe     --JOIN RECIPE INGREDIENTS
WHERE 
ri.id_ingredient IN (1,2)                                           --EGGS OR MILK
AND r.id NOT IN (SELECT ri2.id FROM recipes_ingredients ri2 WHERE ri2.id NOT IN (1,2)) --AND JUST EGGSD OR MILK
HAVING COUNT(ri.Id) = 2                                             --AND THERE ARE 2 INGREDIENTS
UNION ALL
SELECT r.Id, COUNT(ri.Id) FROM recipes r                            --GET THE RECIPE ID AND COUNT OF INGREDIENTS
INNER JOIN recipes_ingredients ri ON r.id_recipe = ri.id_recipe     --JOIN RECIPE INGREDIENTS
WHERE 
ri.id_ingredient = 1                                                --EGGS
AND r.id NOT IN (SELECT ri2.id FROM recipes_ingredients ri2 WHERE ri2.id !=1)  --AND JUST EGGS
HAVING COUNT(ri.Id) = 1     

Answer 3

查询各个成分，并通过将查询中的成分数量与食谱的成分数量相匹配来确保不再有。

编辑

需要的成分一一匹配，以确保所有这些成分都存在于相关食谱中。如果它们都存在并且成分的数量匹配，则不会有更多或更少的成分。

select * 
from recipe_table rt
where rt.id_recipe in (select rig.id_recipe
                    from recipe_ingredient_table rig
                    where rig.id_recipe = rt.id_recipe
                      and <ing_id_1> in (select rig2.id_ingredient from recipe_ingredient_table rig2 where rig2.id_recipe = rig.id_recipe)
                      and <ing_id_2> in (select rig2.id_ingredient from recipe_ingredient_table rig2 where rig2.id_recipe = rig.id_recipe)
                      and ...
                      and (select count(*) 
                           from recipe_ingredient_table
                           where id_recipe = rig.id_recipe) = <number of ingredients>
          )

Answer 4

-- 1 egg only
SELECT *
FROM recipes AS r
WHERE EXISTS (
    SELECT ri.id_recipe, count(*)
    FROM recipes_ingredients AS ri
    WHERE ri.id_recipe = id_recipe 
    AND ri.id_ingredient = 1
    GROUP BY ri.id_recipe
    HAVING COUNT(*) = 1
)

-- 1 egg only OR 2 milk only OR egg and milk
SELECT *
FROM recipes AS r
WHERE EXISTS (
    SELECT ri.id_recipe, count(*)
    FROM recipes_ingredients AS ri
    WHERE ri.id_recipe = id_recipe 
    AND ri.id_ingredient IN (1, 2)
    GROUP BY ri.id_recipe
    HAVING COUNT(*) < 3
)

-- both egg and milk only
-- change above SQL: HAVING COUNT(*) = 2