Question

我需要执行查询才能获得孩子们最喜欢的糖果，这就是我所拥有的：

SELECT COUNT(*) as counts,candies.name
FROM candies 
INNER JOIN kid_candy ON kid_candy.candy_id = candies.id
INNER JOIN kids ON kids.id = kid_candy.kid_id
GROUP BY candies.name

会返回：

counts  | name
--------+---------
  3     | snowbear
  3     | whiterabbit
  2     | lollipop

我想看到的只是

counts  | name
--------+---------
  3     | snowbear
  3     | whiterabbit

那我的查询应该是什么？

Answer 1

所以，首先让我们得到每个糖果名称的计数（tmp_table），然后得到所有糖果的最大数量（max_cnt），最后把它们放在一起，从{{1获得糖果来自tmp_table表...

的计数等于max(counts)

max_cnt

Answer 2

假设孩子和糖果之间的经典的n：m关系，如下所示：

How to implement a many-to-many relationship in PostgreSQL?

您应该在问题中提供此类详细信息。

SELECT c.id, kc.counts, c.name
FROM  (
   SELECT candy_id AS id, count(*) AS counts
        , rank() OVER (ORDER BY count(*) DESC) AS rnk
   FROM   kid_candy
   GROUP  BY 1
   ) kc
JOIN   candies c USING (id)
WHERE  kc.rnk = 1;

这应该比具有两个CTE和不必要的连接的查询快得多。

重点

按candies.name分组可能是错误的（而且更贵）。该名称可能不是唯一的。使用主键列，可能是candies.id
假设参照完整性，我们根本不需要加入表kids。
由于我们需要检查整个表格，因此首先聚合和加入到糖果以获取name 更快后
您可以在聚合函数上运行窗口函数：
- Get the distinct sum of a joined table column

Answer 3

试试这个

SELECT * FROM
(
   SELECT COUNT(*) AS Counts,candies.name FROM candies 
     INNER JOIN kid_candy ON kid_candy.candy_id = candies.id
     INNER JOIN kids ON kids.id = kid_candy.kid_id
   GROUP BY candies.name
 ) T1 JOIN
SELECT MAX(counts) MaxCount FROM
(
   SELECT COUNT(*) AS Counts,candies.name FROM candies 
      INNER JOIN kid_candy ON kid_candy.candy_id = candies.id
      INNER JOIN kids ON kids.id = kid_candy.kid_id
   GROUP BY candies.name
) T1 ON T1.counts =T2.MaxCount

Answer 4

进行子选择

Select max(counts) 
    (select COUNT(*) as counts,candies.name from candies 
INNER JOIN kid_candy ON kid_candy.candy_id = candies.id
INNER JOIN kids ON kids.id = kid_candy.kid_id
GROUP BY candies.name)

获取具有最大计数的行

4 个答案:

重点