Question

我有一个如下所示的表格，想要识别重复的推荐

Client      Team                Referred Date

client1     Referred Team1      2016-02-16
client1     Referred Team1      2016-02-16
client1     Referred Team1      2016-02-16
client1     Referred Team1      2016-01-28
Client2     Referred Team4      2015-07-03
Client2     Referred Team4      2015-07-03
Client3     Referred Team7      2015-04-09
Client3     Referred Team7      2015-04-09
Client3     Referred Team7      2015-04-09
Client3     Referred Team2      2016-09-28
Client3     Referred Team1      2016-10-20
Client4     Referred Team8      2016-11-18

我的查询到目前为止，但我似乎没有得到理想的结果

SELECT 
ClientId
,Team
,COUNT(*) as DuplicateCount

FROM MyData
group by 
ClientId
,Team
having COUNT(*) >1

我想要以下结果

Client      Team                Duplicate Count (Times referred to the same team)

client1     Referred Team1      4
Client2     Referred Team4      2
Client3     Referred Team7      3
Client4     Referred Team8      1

提前致谢

Answer 1

您似乎希望团队中每个客户端的行数最多。这在统计信息中称为模式：

SELECT ct.*
FROM (SELECT ClientId, Team, COUNT(*) as DuplicateCount,
             ROW_NUMBER() OVER (PARTITION BY ClientId ORDER BY COUNT(*) DESC) as seqnum
      FROM MyData
      GROUP BY ClientId, Team
     ) ct
WHERE seqnum = 1;

您可以使用HAVING或WHERE子句过滤掉非重复项。

编辑：

Kannan的答案稍有不同，因此无需使用子查询：

SELECT TOP (1) WITH TIES ClientId, Team, COUNT(*) as DuplicateCount
FROM MyData
GROUP BY ClientId, Team
ORDER BY ROW_NUMBER() OVER (PARTITION BY ClientId ORDER BY COUNT(*) DESC)

Answer 2

您可以使用row_number和子查询，如下所示

 Select top (1) with ties * from (
     Select Client, Team , Cnt = Count(ReferredDate)
         from yourtable 
         group by Client, Team ) a
     order by row_number() over(partition by Client order by cnt desc)

Answer 3

;With cte(Client,Team ,ReferredDate)
AS
(
SELECT 'Client1','Referred Team1','2016-02-16' Union all
SELECT 'Client1','Referred Team1','2016-02-16' Union all
SELECT 'Client1','Referred Team1','2016-02-16' Union all
SELECT 'Client1','Referred Team1','2016-01-28' Union all
SELECT 'Client2','Referred Team4','2015-07-03' Union all
SELECT 'Client2','Referred Team4','2015-07-03' Union all
SELECT 'Client3','Referred Team7','2015-04-09' Union all
SELECT 'Client3','Referred Team7','2015-04-09' Union all
SELECT 'Client3','Referred Team7','2015-04-09' Union all
SELECT 'Client3','Referred Team2','2016-09-28' Union all
SELECT 'Client3','Referred Team1','2016-10-20' Union all
SELECT 'Client4','Referred Team8','2016-11-18' 
)
SELECT Client
    ,Team
    ,DupilcateTeamCount
FROM (
    SELECT Client
        ,Team
        ,DupilcateTeamCount
        ,ROW_NUMBER() OVER (PARTITION BY Client ORDER BY Client ) AS Seq
    FROM (
        SELECT Client
            ,Team
            ,DupilcateTeamCount
            ,ROW_NUMBER() OVER (PARTITION BY Team ORDER BY Client) CCount
        FROM (
            SELECT *,COunt(Team) OVER (PARTITION BY Client,Team ORDER BY Team) AS DupilcateTeamCount
            FROM cte
            ) Dt
        ) DT2
    WHERE DT2.CCount = 1
    ) final
WHERE final.Seq = 1

输出

Client  Team            DupilcateTeamCount
----------------------------------------
Client1 Referred Team1       4
Client2 Referred Team4       2
Client3 Referred Team7       3
Client4 Referred Team8       1

Answer 4

我猜你需要的是将每个客户/团队对的不同引用日期统计为重复。可以通过row_number（）函数跟踪。

查看以下查询是否有效：

drop table test purge;
create table test (Client Varchar2(20), 
                       Team   Varchar2(20), 
                       ReferredDate Date);

insert into test 
select * from (

SELECT 'Client1','Referred Team1',to_date('2016-02-16','YYYY-MM-DD') from dual Union all
SELECT 'Client1','Referred Team1',to_date('2016-02-16','YYYY-MM-DD') from dual Union all
SELECT 'Client1','Referred Team1',to_date('2016-02-16','YYYY-MM-DD') from dual Union all
SELECT 'Client1','Referred Team1',to_date('2016-01-28','YYYY-MM-DD') from dual Union all
SELECT 'Client2','Referred Team4',to_date('2015-07-03','YYYY-MM-DD') from dual Union all
SELECT 'Client2','Referred Team4',to_date('2015-07-03','YYYY-MM-DD') from dual Union all
SELECT 'Client3','Referred Team7',to_date('2015-04-09','YYYY-MM-DD') from dual Union all
SELECT 'Client3','Referred Team7',to_date('2015-04-09','YYYY-MM-DD') from dual Union all
SELECT 'Client3','Referred Team7',to_date('2015-04-09','YYYY-MM-DD') from dual Union all
SELECT 'Client3','Referred Team2',to_date('2016-09-28','YYYY-MM-DD') from dual Union all
SELECT 'Client3','Referred Team1',to_date('2016-10-20','YYYY-MM-DD') from dual Union all
SELECT 'Client4','Referred Team8',to_date('2016-11-18','YYYY-MM-DD') from dual

);

commit;

---=========================================================================================
with t1 as (
select client, team, referreddate, row_number() over (partition by client, team order by referreddate) as dup_cnt
from test 
     )
select distinct client, team, max(dup_cnt)
from t1
group by client, team
order by 1,2
;

输出应为：

CLIENT  TEAM    MAX(DUP_CNT)
1   Client1 Referred Team1  4
2   Client2 Referred Team4  2
3   Client3 Referred Team1  1
4   Client3 Referred Team2  1
5   Client3 Referred Team7  3
6   Client4 Referred Team8  1

检查列是否具有相同的客户端组值

4 个答案: