Question

我有一张桌子

ID   | khk | ts
11AH | 10   | 2
11AH | 10   | 2
11AH | 9    | 1
22AH | 5    | 2
22AH | 5    | 3

我需要分组ID的sum（ts）（但仅当计数id大于2时才需要），并且仅当分组的khk计数大于1时才从中求和）

所以对于id 11AH，sum（ts）= 4（而不是5），对于22AH，什么都不是，因为它们只有两个。

我尝试这样的方法，但是不正确

select sum(ts),id from table group by id,khk having count(id)>2 and count(khk)>1;

Answer 1

尝试以下

DEMO

select * from
(select id,khk, sum(ts) as ts,(select count(id) from cte1 b where a.id=b.id) as cnt
from cte1 a 
group by id,khk 
having count(*)>1)X where cnt>2

输出：

id      khk  ts  cnt
11AH    10   4   3

Answer 2

您可以尝试使用窗口功能

with cte
(
select id,khk,ts, count(*) over(partition by id order by id) as id_cnt,
count(*) over(partition by khk order by khk) as khk_cnt
) select id,khk,sum(ts)
 where id_cnt>2 and khk_cnt>1
 group by id,khk

Answer 3

我建议使用分析功能而不是子查询：

select id, khk, sum_ts
from (select t.id, t.khk, sum(ts) as sum_ts,
             count(*) as cnt_id_khk, 
             count(*) over (partition by t.id) as cnt_id
      from t
      group by id, khk 
     ) t
where cnt_id_khk > 1 and cnt_id > 1

基于两个分组的SQL和

3 个答案: