Question

我有一张桌子：

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

选择第二和第三高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 |
+---------+------------+--------------+

Answer 1

; 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

Sql Fiddle is here

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

Sql Fiddle for this is here

Answer 2

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

Answer 3

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

为组选择第2和第3个最高数字

3 个答案: