Question

鉴于以下表格：

Recipes
| id | name
| 1  | 'chocolate cream pie'
| 2  | 'banana cream pie'
| 3  | 'chocolate banana surprise'

Ingredients
| id | name
| 1  | 'banana'
| 2  | 'cream'
| 3  | 'chocolate'

RecipeIngredients
| recipe_id | ingredient_id
|     1     |      2
|     1     |      3
|     2     |      1
|     2     |      2
|     3     |      1
|     3     |      3

如何构建SQL查询以查找recipe.name ='chocolate'和ingredients.name ='cream'的食谱？

Answer 1

使用：

  SELECT r.name
    FROM RECIPES r
    JOIN RECIPEINGREDIENTS ri ON ri.recipe_id = r.id
    JOIN INGREDIENTS i ON i.id = ri.ingredient_id
                      AND i.name IN ('chocolate', 'cream')
GROUP BY r.name
  HAVING COUNT(DISTINCT i.name) = 2

这里的关键点是计数必须等于成分名称的数量。如果它不是一个独特的计数，则由于重复而存在误报的风险。

Answer 2

这称为关系师。讨论了各种技术here。

尚未给出的另一种选择是双重不存在

SELECT r.id, r.name
FROM Recipes r
WHERE NOT EXISTS (SELECT * FROM Ingredients i
                  WHERE name IN ('chocolate', 'cream')
                  AND NOT EXISTS
                      (SELECT * FROM RecipeIngredients ri
                       WHERE ri.recipe_id = r.id
                       AND ri.ingredient_id = i.id))

Answer 3

如果您正在搜索多个关联，那么编写查询的最简单方法是使用多个EXISTS条件而不是单个JOIN。

SELECT r.id, r.name
FROM Recipes r
WHERE EXISTS
(
    SELECT 1
    FROM RecipeIngredients ri
    INNER JOIN Ingredients i
        ON i.id = ri.ingredient_id
    WHERE ri.recipe_id = r.id
    AND i.name = 'chocolate'
)
AND EXISTS
(
    SELECT 1
    FROM RecipeIngredients ri
    INNER JOIN Ingredients i
        ON i.id = ri.ingredient_id
    WHERE ri.recipe_id = r.id
    AND i.name = 'cream'
)

如果您确定关联是唯一的（即单个配方只能包含每个成分的单个实例），那么您可以使用具有COUNT函数和可能速度的分组子查询作弊它（性能将取决于DBMS）：

SELECT r.id, r.Name
FROM Recipes r
INNER JOIN RecipeIngredients ri
    ON ri.recipe_id = r.id
INNER JOIN Ingredients i
    ON i.id = ri.ingredient_id
WHERE i.name IN ('chocolate', 'cream')
GROUP BY r.id, r.Name
HAVING COUNT(*) = 2

或者，如果某个配方可能有多个相同成分的实例（UNIQUE关联表上没有RecipeIngredients约束），则可以将最后一行替换为：

HAVING COUNT(DISTINCT i.name) = 2

Answer 4

select r.*
from Recipes r
inner join (
    select ri.recipe_id
    from RecipeIngredients ri 
    inner join Ingredients i on ri.ingredient_id = i.id
    where i.name in ('chocolate', 'cream')
    group by ri.recipe_id
    having count(distinct ri.ingredient_id) = 2
) rm on r.id = rm.recipe_id

Answer 5

SELECT DISTINCT r.id, r.name
FROM Recipes r
INNER JOIN RecipeIngredients ri ON
    ri.recipe_id = r.id
INNER JOIN Ingredients i ON
    i.id = ri.ingredient_id
WHERE
    i.name IN ( 'cream', 'chocolate' )

编辑以下评论，谢谢！这是正确的方式：

SELECT DISTINCT r.id, r.name
FROM Recipes r
INNER JOIN RecipeIngredients ri ON
    ri.recipe_id = r.id
INNER JOIN Ingredients i ON
    i.id = ri.ingredient_id AND
    i.name = 'cream'
INNER JOIN Ingredients i2 ON
    i2.id = ri.ingredient_id AND
    i2.name = 'chocolate'

Answer 6

另一种方式：

版本2（作为存储过程）修订

select   r.name
from   recipes r
where   r.id  = (select  t1.recipe_id
        from  RecipeIngredients t1 inner join
     RecipeIngredients     t2 on t1.recipe_id = t2.recipe_id
     and     t1.ingredient_id = @recipeId1
     and     t2.ingredient_id = @recipeId2)

编辑2： [在人们开始尖叫之前]：）

这可以放在版本2的顶部，这将允许按名称查询而不是传入id。

select @recipeId1 = recipe_id from Ingredients where name = @Ingredient1
select @recipeId2 = recipe_id from Ingredients where name = @Ingredient2

我已经测试了第2版，但它确实有效。大多数用户在成分表上链接，在这种情况下完全不需要！

编辑3 :(测试结果）;

运行此存储过程时，这些是结果。

结果具有格式（First Recipe_id; Second Recipe_id，Result）

1,1, Failed
1,2, 'banana cream pie'
1,3, 'chocolate banana surprise'
2,1, 'banana cream pie'
2,2, Failed
2,3, 'chocolate cream pie'
3,1, 'chocolate banana surprise'
3,2, 'chocolate cream pie'
3,3, Failed

显然，当两个约束相同时，此查询不处理大小写，但适用于所有其他情况。

编辑4 :(处理相同的约束案例）：

替换此行：

r.id = (select t1...

到

r.id in (select t1...

使用失败的案例来提供：

1,1, 'banana cream pie' and 'chocolate banana surprise'
2,2, 'chocolate cream pie' and 'banana cream pie'
3,3, 'chocolate cream pie' and 'chocolate banana surprise'

SQL查询通过中间表

6 个答案: