Question

表

recipe_food_xref
recipe_id int
food_id int

需要在recipe_food_xref中找到一条记录，其中食谱只有一种食物，食物是特定食物。

它与自身完美结合：

SELECT x1.recipe_id FROM recipe_food_xref x1
INNER JOIN recipe_food_xref x2 ON x2.recipe_id = x1.recipe_id
WHERE x1.food_id = 1
GROUP BY x1.recipe_id
HAVING COUNT(x2.recipe_id) = 1

这看起来有点难看，我想知道是否有更好的方法。

这是一个包含一些示例数据的SqlFiddle。基本上我想找到recipe_id：1，因为它有food_id：1并且没有多个food_id

http://sqlfiddle.com/#!3/6d474/1

Answer 1

SELECT recipe_id 
FROM recipe_food_xref
GROUP BY recipe_id
HAVING sum(case when food_id = 1 then 1 else 0 end) = 1
and sum(case when food_id <> 1 then 1 else 0 end) = 0

SQLFiddle demo

Answer 2

这是我第一次去：

SELECT recipe_id
FROM recipe_food_xref
GROUP BY recipe_id
HAVING COUNT(food_id) = 1 AND SUM(food_id) = 1;

请注意，常规格式为HAVING COUNT(FOOD_ID) = 1 AND SUM(FOOD_ID) = <food id>。这是有效的，因为如果只有一个food_id，则总和等于id。

似乎与Juergen的答案类似，但更简单。我错过了什么吗？

我还尝试了以下操作，它必须额外扫描表格，但因为我之前从未使用ALL关键字，所以我觉得它很有趣。

SELECT recipe_id
FROM recipe_food_xref AS r
WHERE 1 = ALL (SELECT food_id FROM recipe_food_xref WHERE r.recipe_id=recipe_id);

http://sqlfiddle.com/#!3/6d474/26

Answer 3

select *
from recipe_food_xref x
where not exists (
                   select i.food_id
                   from recipe_food_xref i
                   where i.recipe_id = x.recipe_id and
                         i.food_id <> x.food_id
                 ) 
-- if this is omitted you get all recipes having just one food:
      and x.food_id = 1

Answer 4

以下是我对问题的解释：

查找所有含有单一成分的食谱。在这些配方中，找到单一成分为X
的配方

SELECT recipe_id
     , Max(food_id) As food_id
     , Count(*) As number_of_ingredients
FROM   recipe_food_xref
GROUP
    BY recipe_id
HAVING Count(*) = 1
AND    Max(food_id) = 1

在连接表中查找只有一行的记录

4 个答案:

SQLFiddle demo