SQL 中的复杂 RANK

Question

我有一个复杂的查询，它真的超出了我的头脑。我认为需要 RANK 的 RANK，但必须有更好的现有方法。

这里我有一个简单的table:

Manufacturer    DateOF  Status              Prefer
Dell            05-2014 ComputerInstalled   30
Dell            05-2014 ComputerUninstalled 70
Dell            05-2014 ComputerUninstalled 70
Dell            05-2014 ComputerUninstalled 70
Dell            05-2014 ComputerInstalled   30
Dell            05-2014 ComputerUninstalled 70
Dell            05-2014 ComputerNew         26
Dell            05-2014 ComputerNew         26
Dell            05-2014 ComputerInstalled   30
Dell            05-2014 ComputerInstalled   30

我需要做的是按 MANUFACTURER 和 DATEOF 列对 table 进行 GROUP BY，然后选择具有最低 PREFER 编号的行（在本例中为 26）。

使用 RANK 函数很容易：

SELECT sq.*
FROM
(
SELECT
*,
RANK() OVER (PARTITION BY Manufacturer,DateOF ORDER BY Prefer) AS RankPrefer
FROM
table1
WHERE
RankPrefer = 1
) sq

所以我将得到状态为 ComputerNew 的 2 行的结果。

Manufacturer    DateOF  Status              Prefer
Dell            05-2014 ComputerNew         26
Dell            05-2014 ComputerNew         26

这很简单，不是问题。

问题是：

我必须执行以下规则：

如果具有最低 Prefer 值的行（例如：26）结果在他们的 Status 字段中有 ComputerNew 值，然后我必须包含更多具有 ComputerInstalled 值的行。

结果应该是这样的：

Manufacturer    DateOF  Status              Prefer
Dell            05-2014 ComputerInstalled   30
Dell            05-2014 ComputerInstalled   30
Dell            05-2014 ComputerNew         26
Dell            05-2014 ComputerNew         26
Dell            05-2014 ComputerInstalled   30
Dell            05-2014 ComputerInstalled   30

类似这个规则，我还有一个：

如果具有最低 Prefer 值的行（例如：26）结果在他们的 Status 字段中有 ComputerOld 值，然后我必须包含更多具有 ComputerUninstalled 值的行。

我认为 RANK of RANKING 可以解决这个问题，但现在我真的迷路了。

如有任何帮助，我们将不胜感激。

谢谢

编辑 1:

Gordon 的解决方案几乎不错，但并不完美。

我给你更多的测试数据，你可以看到哪里失败了。要测试的 SQLFiddle 是 here.

我把测试数据也放在这里：

INSERT Table1 VALUES ('HP10011','04/01/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP10011','04/04/2014','ComputerOld',26)
INSERT Table1 VALUES ('HP10011','04/04/2014','ComputerOld',26)
INSERT Table1 VALUES ('HP10011','04/30/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP10011','05/23/2014','QuickDispose',10)
INSERT Table1 VALUES ('HP10011','06/03/2014','QuickDispose',10)
INSERT Table1 VALUES ('HP10077','04/01/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP1910','04/25/2014','QuickDispose',10)
INSERT Table1 VALUES ('HP1910','05/01/2014','ComputerInstalled',30)
INSERT Table1 VALUES ('HP1910','05/01/2014','ComputerInstalled',30)
INSERT Table1 VALUES ('HP1910','05/01/2014','ComputerInstalled',30)
INSERT Table1 VALUES ('HP1910','05/01/2014','ComputerInstalled',30)
INSERT Table1 VALUES ('HP1910','05/01/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP1910','05/01/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP1910','05/01/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP1910','05/01/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP1910','05/02/2014','ComputerInstalled',30)
INSERT Table1 VALUES ('HP1910','05/02/2014','ComputerInstalled',30)
INSERT Table1 VALUES ('HP3720','05/07/2014','ComputerInstalled',30)
INSERT Table1 VALUES ('HP3720','05/07/2014','ComputerInstalled',30)
INSERT Table1 VALUES ('HP3720','05/07/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP3720','05/07/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP3720','05/07/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP3720','05/07/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP3720','05/08/2014','ComputerInstalled',30)
INSERT Table1 VALUES ('HP3720','05/08/2014','ComputerInstalled',30)
INSERT Table1 VALUES ('HP3720','05/08/2014','ComputerInstalled',30)
INSERT Table1 VALUES ('HP3720','05/08/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP3720','06/06/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP3720','06/06/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP3720','06/10/2014','ComputerOld',26)
INSERT Table1 VALUES ('HP3720','06/10/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP3720','06/10/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP3720','06/11/2014','ComputerOld',26)
INSERT Table1 VALUES ('HP3720','06/11/2014','ComputerUninstalled',70)
INSERT Table1 VALUES ('HP3720','06/11/2014','ComputerUninstalled',70)

查询 returns ComputerInstalled 和 ComputerUninstalled 两行对于以下数据：

'HP1910','05/01/2014','ComputerInstalled',30
'HP1910','05/01/2014','ComputerUninstalled',70

它应该只选择 ComputerInstalled，因为对于那个 Manufacturer，在同一个月，它应该选择最低的 Prefer (30)。

这个数据集的结果应该是这样的：

Manufacturer    DateOF  Status          Prefer
HP10011 2014-04-01  ComputerUninstalled 70
HP10011 2014-04-04  ComputerOld         26
HP10011 2014-04-04  ComputerOld         26
HP10011 2014-04-30  ComputerUninstalled 70
HP10011 2014-05-23  QuickDispose        10
HP10011 2014-06-03  QuickDispose        10
HP10077 2014-04-01  ComputerUninstalled 70
HP1910  2014-04-25  QuickDispose        10
HP1910  2014-05-01  ComputerInstalled   30
HP1910  2014-05-01  ComputerInstalled   30
HP1910  2014-05-01  ComputerInstalled   30
HP1910  2014-05-01  ComputerInstalled   30
HP3720  2014-05-07  ComputerInstalled   30
HP3720  2014-05-07  ComputerInstalled   30
HP3720  2014-05-08  ComputerInstalled   30
HP3720  2014-05-08  ComputerInstalled   30
HP3720  2014-05-08  ComputerInstalled   30
HP3720  2014-06-06  ComputerUninstalled 70
HP3720  2014-06-06  ComputerUninstalled 70
HP3720  2014-06-10  ComputerOld         26
HP3720  2014-06-10  ComputerUninstalled 70
HP3720  2014-06-10  ComputerUninstalled 70
HP3720  2014-06-11  ComputerOld         26
HP3720  2014-06-11  ComputerUninstalled 70
HP3720  2014-06-11  ComputerUninstalled 70

Answer 1

我认为这应该能满足您的需求：

WITH sq AS
( SELECT *, RANK() OVER (PARTITION BY Manufacturer,DateOF ORDER BY Prefer) AS RankPrefer
  FROM table1
)
SELECT *
FROM   sq
WHERE  RankPrefer <= (SELECT TOP 1 RankPrefer FROM sq WHERE Status != 'ComputerNew' ORDER BY RankPrefer)

Here is the SqlFiddle.

Answer 2

这是一个想法。找出行的偏好排名。然后使用 exists.

确定 rank = 1 的行是否符合您的条件

最终查询如下所示：

with r as (
      select t.*,
             rank() over (partition by manufacturer, dateof order by Prefer) as seqnum
      from table1 t
     ),
     r1 as (
      select r.*
      from r
      where seqnum = 1
     )
select r.*
from r
where r.seqnum = 1 or
      (exists (select 1 from r1 where status = 'ComputerNew' and r1.dateof = r.dateof) and r.status = 'ComputerInstalled' or
       exists (select 1 from r1 where status = 'ComputerOld' and r1.dateof = r.dateof) and r.status = 'ComputerUninstalled'
      );

Answer 3

好的，既然您已经对问题进行了一些编辑，我有一个不同的答案，我相信可以解决问题。这是查询：

;with r as (
      select t.*,
             CAST(MONTH(dateof) AS VARCHAR(2)) + '-' + CAST(YEAR(dateof) AS VARCHAR(4)) AS EffDate,
             rank() over (partition by manufacturer, CAST(MONTH(dateof) AS VARCHAR(2)) + '-' + CAST(YEAR(dateof) AS VARCHAR(4)) order by Prefer) as seqnum
      from Table1 t
     ),
     r1 as (
      select r.*
      from r
      where seqnum = 1
     )
select r.*
from r
where r.seqnum = 1 
or
(
  r.Status = 'ComputerUninstalled' and 
  exists ( Select 1 
           from  r1 
           where r1.Manufacturer = r.Manufacturer
           and   r1.EffDate = r.EffDate
           and   r1.Status = 'ComputerOld' )
  and   r.seqNum = ( Select Min(SeqNum) From r as r2
                      Where r2.Manufacturer = r.Manufacturer
                      And   r2.EffDate = r.EffDate
                      And   r2.SeqNum > 1 )          
)
or
(
  r.Status = 'ComputerInstalled' and 
  exists ( Select 1 
           from  r1 
           where r1.Manufacturer = r.Manufacturer
           and   r1.EffDate = r.EffDate
           and   r1.Status = 'ComputerNew' )
  and   r.seqNum = ( Select Min(SeqNum) From r as r2
                      Where r2.Manufacturer = r.Manufacturer
                      And   r2.EffDate = r.EffDate
                      And   r2.SeqNum > 1 )          
);

注意：我得到的记录比您预期的结果集显示的多 2 条。但是根据您的描述，我认为您的预期结果有误。 2014 年 5 月有 6 个 "ComputerInstalled" 的 HP1910，Prefer 为 30。其中 4 个日期为 5 月 1 日，2 个日期为 5 月 2 日。你遗漏了 5 月 2 日的记录。除此之外，这个结果集符合您的预期结果，我相信应该适用于更大的数据集。

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