Question

我正在练习一些SQL，我想到了以下问题：

对于每个酒吧，找到更多人去的时间。我有以下表格：

            GOESTO
id_person   id_pub  time
1           1       Daytime
2           2       Night time
3           3       All Day
4           1       Daytime
5           2       Night time
6           1       All Day
7           3       Daytime
8           3       Night time
9           3       Night time
10          1       Night time

         PUB
id_pub  pub_name    cost
1       pub1        123
2       pub2        324
3       pub3        345

我想得的是以下内容：

pub_name   time

我认为我应该使用MAX和COUNT函数，但我不确定应该怎么做。它应该在Oracle数据库中工作。

谢谢！

Answer 1

试试这个：

WITH mydata AS (
      select id_pub, "TIME", count(*) as cnt
      from GOESTO
      group by id_pub, "TIME"
) 
SELECT m.id_pub, m."TIME", m.cnt
FROM mydata m
JOIN (
   SELECT id_pub, max( cnt ) as cnt
   FROM mydata
   GROUP BY  id_pub
) x
ON (m.id_pub = x.id_pub AND m.cnt = x.cnt);

或者这个

SELECT id_pub, "TIME"
FROM (
    SELECT t.*,
           dense_rank() over (partition by id_pub order by cnt desc ) rnk
    FROM (
          select id_pub, "TIME", count(*) as cnt
          from GOESTO
          group by id_pub, "TIME"
    ) t
)
WHERE rnk = 1

要获取名称而不是id_pub值，您需要使用PUB表

加入上述查询

SELECT p.pub_name, q."TIME"
FROM (  one_of_the_above_query )q
JOIN PUB p
ON p.id_pub = q.id_pub

查找每个关联属性的最重复值

1 个答案: