CREATE TABLE [dbo].[TargettingReport](
[MLISNPI] [varchar](50) NULL,
[CLIENT_ID1] [varchar](50) NULL,
[Sum of count] [int] NULL
) ON [PRIMARY]
我正在尝试根据SUM OF COUNT
MLISNPI, CLIENT_ID1
这就是我获得MLISNPI, CLIENT_ID1 每次出现的最大值的方式：
select [MLISNPI],[CLIENT_ID1],MAX([Sum of count])
from [TargettingReport]
group by
[MLISNPI],[CLIENT_ID1]
问题是如何获得第二和第三高？
这是一些示例数据：
+---------+------------+--------------+
| MLISNPI | CLIENT_ID1 | sum of count |
+---------+------------+--------------+
| 567890 | 214060 | 18 |
| 678901 | 214060 | 58 |
| 789012 | 214060 | 27 |
| 891012 | 214060 | 1 |
| 101112 | 214060 | 23 |
| 003001 | 101596 | 0 |
| 003001 | 101522 | 436 |
| 003001 | 101597 | 0 |
| 003002 | 102165 | 66 |
| 003002 | 100062 | 1 |
| 003002 | 211074 | 1229 |
| 003006 | 102235 | 21 |
| 003014 | 213926 | 5 |
| 003016 | 213143 | 3 |
| 003023 | 213801 | 55 |
| 003023 | 212876 | 44 |
| 003023 | 100218 | 0 |
| 003028 | 211144 | 133 |
| 003041 | 100236 | 346 |
| 003041 | 103164 | 65 |
| 003051 | 213402 | 157 |
| 003058 | 100572 | 28 |
| 003065 | 101632 | 29 |
| 003071 | 213632 | 6 |
| 003072 | 101506 | 4 |
| 003081 | 100087 | 398 |
| 003083 | 214311 | 7 |
| 003117 | 210178 | 203 |
| 003121 | 214008 | 9 |
| 003139 | 102179 | 1635 |
| 003147 | 216022 | 21 |
| 003149 | 211425 | 1 |
| 003186 | 215748 | 1 |
+---------+------------+--------------+
答案 0 :(得分：4)
; with numbered as
(
select *,
-- Assign number to each record
-- Numbering is separate for each pair of MLISNPI and CLIENT_ID1
-- Highest [Sum of desc] will get number 1 and so forth
row_number() over (partition by [MLISNPI], [CLIENT_ID1]
order by [Sum of count] desc) rn
from [TargettingReport]
)
select [MLISNPI],
[CLIENT_ID1],
[Sum of count]
from numbered
-- Now one can filter ranks
where rn between 2 and 3
更新：在mlisnpi周围旋转[计数总和]
; with numbered as
(
select mlisnpi,
[sum of count],
-- Assign number to each record
-- Numbering is separate for each MLISNPI
-- Highest [Sum of desc] will get number 1 and so forth
row_number() over (partition by [MLISNPI]
order by [Sum of count] desc) rn
from [TargettingReport]
)
select mlisnpi,
[1] Rank1,
[2] Rank2,
[3] Rank3
from numbered
pivot (max([Sum of count]) for rn in ([1], [2], [3])) p
如果您确实需要Cliend_ID1而不是[Sum of count]，只需将所有出现的[Sum of count]替换为Client_ID1。不要将Client_ID1添加到cte，因为pivot会将其包含在结果中，并在其自己的行中显示mlisnpi和Client_ID1的每个组合。
UPDATE2 ：包含标题的旋转。更加冗长，因为您需要联合所有标题（client_id1）并操纵行号以获得正确的标题和总和组合。您还需要将所有数据转换为相同的数据类型（此处为varchar）才能使用union all。
; with numbered as
(
select *,
row_number() over (partition by [MLISNPI]
order by [Sum of count] desc) rn
from [TargettingReport]
),
unioned as
(
select mlisnpi,
convert(varchar(20), [Sum of count]) value,
rn * 2 rn
from numbered
union all
select mlisnpi,
convert(varchar(20), [Client_ID1]),
rn * 2 - 1
from numbered
)
select mlisnpi,
[1] Client1,
[2] Rank1,
[3] Client2,
[4] Rank2,
[5] Client3,
[6] Rank3
from unioned
pivot (max(value) for rn in ([1], [2], [3], [4], [5], [6])) p
答案 1 :(得分：1)
创建3个顶部值的视图
CREATE VIEW TESTVIEW
AS
SELECT ABB4.[MLISNPI], ABB4.[CLIENT_ID1], MAX(ABB4.[SUM OF COUNT]) AS NEW3
FROM TARGETTINGREPORT AS ABB4
LEFT JOIN (SELECT ABB2.[MLISNPI], ABB2.[CLIENT_ID1], MAX([SUM OF COUNT]) AS NEW2
FROM TARGETTINGREPORT AS ABB2
LEFT JOIN (SELECT [MLISNPI], [CLIENT_ID1], MAX([SUM OF COUNT]) AS NEW1
FROM TARGETTINGREPORT GROUP BY MLISNPI, [CLIENT_ID1])
AS ABB1 ON ABB1.MLISNPI = ABB2.MLISNPI AND ABB1.NEW1 = ABB2.[SUM OF COUNT] and ABB1.CLIENT_ID1 = ABB2.CLIENT_ID1
WHERE ABB1.MLISNPI IS NULL
GROUP BY ABB2.MLISNPI, ABB2.[CLIENT_ID1])
AS ABB3 ON ABB3.MLISNPI = ABB4.MLISNPI AND ABB3.NEW2 = ABB4.[SUM OF COUNT] AND ABB3.CLIENT_ID1 = ABB4.CLIENT_ID1
LEFT JOIN (SELECT ABB5.[MLISNPI], ABB5.[CLIENT_ID1], MAX([SUM OF COUNT]) AS NEW4
FROM TARGETTINGREPORT AS ABB5
GROUP BY ABB5.MLISNPI, ABB5.[CLIENT_ID1])
AS ABB6 ON ABB6.MLISNPI = ABB4.MLISNPI AND ABB6.NEW4 = ABB4.[SUM OF COUNT] AND ABB6.CLIENT_ID1 = ABB4.CLIENT_ID1
WHERE ABB3.MLISNPI IS NULL AND ABB6.MLISNPI IS NULL
GROUP BY ABB4.MLISNPI, ABB4.[CLIENT_ID1]
UNION ALL
SELECT ABB2.[MLISNPI], ABB2.[CLIENT_ID1], MAX([SUM OF COUNT]) AS NEW2
FROM TARGETTINGREPORT AS ABB2
LEFT JOIN (SELECT [MLISNPI], [CLIENT_ID1], MAX([SUM OF COUNT]) AS NEW1
FROM TARGETTINGREPORT GROUP BY MLISNPI, [CLIENT_ID1])
AS ABB1 ON ABB1.MLISNPI = ABB2.MLISNPI AND ABB1.NEW1 = ABB2.[SUM OF COUNT] AND ABB1.CLIENT_ID1 = ABB2.CLIENT_ID1
WHERE ABB1.MLISNPI IS NULL
GROUP BY ABB2.MLISNPI, ABB2.[CLIENT_ID1]
UNION ALL
SELECT [MLISNPI], [CLIENT_ID1], MAX([SUM OF COUNT]) AS NEW1
FROM TARGETTINGREPORT
GROUP BY MLISNPI, [CLIENT_ID1]
然后创建另一个视图
CREATE VIEW TESTVIEW2
AS
SELECT ABB1.MLISNPI, ABB1.CLIENT_ID1, ABB1.NEW3
FROM TESTVIEW AS ABB1
LEFT JOIN (SELECT MLISNPI, CLIENT_ID1, MIN(NEW3) AS VAR1
FROM TESTVIEW
GROUP BY MLISNPI, CLIENT_ID1) AS ABB2
ON ABB2.MLISNPI = ABB1.MLISNPI AND ABB2.CLIENT_ID1 = ABB1.CLIENT_ID1 AND ABB2.VAR1 = ABB1.NEW3
LEFT JOIN (SELECT MLISNPI, CLIENT_ID1, MAX(NEW3) AS VAR2
FROM TESTVIEW
GROUP BY MLISNPI, CLIENT_ID1) AS ABB3
ON ABB3.MLISNPI = ABB1.MLISNPI AND ABB3.CLIENT_ID1 = ABB1.CLIENT_ID1 AND ABB3.VAR2 = ABB1.NEW3
WHERE ABB2.MLISNPI IS NULL AND ABB3.MLISNPI IS NULL
最后以数据透视表格形式调用3个最高值
SELECT TESTVIEW.MLISNPI, TESTVIEW.CLIENT_ID1, MAX(TESTVIEW.NEW3) AS '1', TESTVIEW2.NEW3 AS '2', MIN(TESTVIEW.NEW3) AS '3'
FROM TESTVIEW
LEFT JOIN TESTVIEW2 ON TESTVIEW2.MLISNPI = TESTVIEW.MLISNPI AND TESTVIEW2.CLIENT_ID1 = TESTVIEW.CLIENT_ID1
GROUP BY TESTVIEW.MLISNPI, TESTVIEW.CLIENT_ID1, TESTVIEW2.NEW3
答案 2 :(得分：0)
你能做到rownum＆lt; 3并从您的选择中删除最大值？
select top 3 [MLISNPI],[CLIENT_ID1], Max([Sum of count]) COUNTS
from [TargettingReport]
group by
[MLISNPI],[CLIENT_ID1]
ORDER by [Sum of count] desc