Question

假设我们有桌子：

create table EVENT("id" number, "date" DATE, "value" number);

我希望获得每个选中的 id N 或更多次的所有行。所以：

    id  |    date    | value
   --------------------------
    1   | 2011-01-01 |  100
    1   | 2011-01-02 |  200
    2   | 2011-01-05 |  300
    2   | 2011-03-15 |  800
    3   | 2011-02-01 |  400
    4   | 2011-01-01 |  500
    4   | 2011-04-21 |  600
    4   | 2011-01-01 |  700

和 N == 2我得到除 id = 3 之外的所有行，对于 N == 3我只获得< EM> ID = 4 ...

我使用Oracle但似乎这种类型的查询需要一些新的SQL知识...

Answer 1

SELECT "id",
       "date",
       "value"
FROM   (SELECT EVENT.*,
               COUNT(*) OVER (PARTITION BY "id") AS CNT
        FROM   EVENT)
WHERE  CNT >= 3

SQL Fiddle

Answer 2

SELECT *
FROM Event
WHERE Id IN (SELECT Id 
             FROM Event
             GROUP BY Id 
             HAVING COUNT(*) > N)

编辑：Martin Smith的答案应该是最好的表现，唯一的缺点是你必须列出字段以避免在结果中包含COUNT（）。

Answer 3

select  *
from    Event
where   id in
        (
        select  id
        from    Event
        group by
                id
        having  count(*) > 3 -- For N = 3
        )

Answer 4

你走了：

SELECT *
FROM tmp
WHERE id IN (SELECT id FROM tmp GROUP BY id 
             HAVING COUNT(*) > N)

根据您的条件更新N的值。

SQL Fiddle DEMO

Answer 5

SELECT * FROM Event
INNER JOIN (
    SELECT id, COUNT(*) AS Cnt FROM Event GROUP BY id
) AS C ON Event.id = C.id
WHERE C.Cnt >= 3

Answer 6

select 
  e.* 
from 
  event e, 
  (select e1.id, count(*) as id_num from event e1 group by e1.id) as e2
where
  e.id = e2.id
  and e2.id_num >= 3

Answer 7

SELECT *
FROM   EVENT
GROUP  BY "id"
HAVING ( COUNT("id") > N - 1
         AND COUNT("id") < N + 1 );

SQL查询选择字段值出现N次或更多次的所有行？

7 个答案:

SQL Fiddle DEMO