选择SQL中组的前N个独占值？

Question

在 SQL（最好是 SQL Server）中有没有办法选择一个组中排在其他组之外的前 N ​​条记录？

例如：

DROP TABLE IF EXISTS #DISTANCE

CREATE TABLE #DISTANCE
(
    GNAME VARCHAR(3)
,   CNAME VARCHAR(3)
,   DIST NUMERIC(5,3)
)

INSERT INTO #DISTANCE
VALUES ('E1', 'C1', 1), ('E1','C2',2),
       ('E2', 'C1', 1.5), ('E2','C2',2.5)

如果我按距离 ASC 为每个 ENAME 寻找第一个专有 CNAME，我希望得到这样的输出：

Ename	Cname	Dist
E1	C1	1
E2	C2	2.5

请注意，省略 E1|C2 和 E2|C1，因为它们将是该组排名结果中的第二个值。

我想出了一些 SQL 方法来尝试正确地提取它，但是当我在 ENAME 上添加其他组并且如果我更改我的 Top N 值时，我的工作就会崩溃。

如果我增加复杂性：

TRUNCATE TABLE #DISTNACE

INSERT INTO #DISTANCE
VALUES ('E1', 'C1', 1), ('E1', 'C2', 2), 
       ('E1', 'C3', 3), ('E1', 'C4', 5),
       ('E2', 'C1', 2.5), ('E2', 'C2', 4), 
       ('E2', 'C3', 3.5), ('E2', 'C4', 6),
       ('E3', 'C4', 7), ('E3', 'C5', 6), 
       ('E3', 'C6', 4)

我试图得到的 SQL 输出如下所示：

GNAME	CNAME	DIST
E1	C1	1.000
E1	C2	2.000
E1	C3	3.000
E2	C4	6.000
E3	C6	4.000
E3	C5	6.000

我可以让它在这个特定的实例中工作，使用这个代码：

WITH X AS  
(
    SELECT * 
    --, RNK             = DENSE_RANK() OVER (ORDER BY DIST ASC) 
    , CNAME_RNK_BY_DIST = DENSE_RANK() OVER (PARTITION BY CNAME ORDER BY DIST ASC)
    , CNAME_RNK_BY_DIST = DENSE_RANK() OVER (PARTITION BY CNAME ORDER BY DIST ASC) 
    FROM #DISTANCE
    )
,MINDIST AS ( -- FIRST OCCURANCE OF CNAME VALUE
    SELECT 
        CNAME
    ,   MIN(DIST) MINDIST
    FROM X GROUP BY CNAME
)
            -- SELECT * , CALC = SUM(CNAME_RNK_BY_DIST / 4) OVER (PARTITION BY CNAME ORDER BY DIST ASC) FROM X order by CNAME, DIST
, X2 AS (

    SELECT *, CALC = SUM(FLOOR(CNAME_RNK_BY_DIST / 4)) OVER (PARTITION BY CNAME ORDER BY DIST ASC) FROM X
)
    --SELECT * FROM X2 order by CNAME, dist
, CALC AS (
    SELECT CNAME, MAXINC = MAX(CALC) FROM X2 GROUP BY CNAME
)
    --SELECT * FROM CALC
, FIRST_OCCURANCE_PAIRS AS (
    SELECT A.*
        ,OCCURANCE = RANK() OVER (PARTITION BY CNAME ORDER BY DIST)
    FROM X A
    JOIN MINDIST B ON A.CNAME = B.CNAME AND A.DIST = B.MINDIST
)
    --SELECT * FROM FIRST_OCCURANCE_PAIRS

,ISO AS 
(
    SELECT * fROM FIRST_OCCURANCE_PAIRS WHERE OCCURANCE > 3 
)
--select * from FIRST_OCCURANCE_PAIRS
--  SELECT * FROM ISO
, NEXT_OCCURANCE AS (
    SELECT A.*
    FROM  X AS A
    JOIN CALC ON A.CNAME = CALC.CNAME
    JOIN ISO B ON A.CNAME_RNK_BY_DIST  =  CALC.MAXINC               and A.CNAME = B.CNAME
)
    --select * from NEXT_OCCURANCE
, FRAME AS (

SELECT 
    CNAME
,   CNAME
,   DIST

FROM FIRST_OCCURANCE_PAIRS
--WHERE OCCURANCE <=3
UNION
SELECT
    CNAME
,   CNAME
,   DIST
FROM NEXT_OCCURANCE
)
--select * from FRAME
, FINAL AS (
    SELECT * ,FINALRNK = ROW_NUMBER() OVER (PARTITION BY CNAME ORDER BY DIST)
    FROM
    FRAME
)
    SELECT * FROM FINAL WHERE FINALRNK <4

但随着更多记录的添加，逻辑失败。有没有办法清理这个 SQL 并获得任意数量组合的结果？

Answer 1

在查看了我的问题之后，我认为我没有为我的问题提供足够有用的信息。经过一些调整，我想我能够达到我想要的。我认为我必须运行批处理执行步骤来传递我的数据以提取我需要的东西，而不是能够在单个查询中执行此操作。也许我有办法减少/优化下面的代码？

DROP TABLE IF EXISTS #DISTANCE
CREATE TABLE #DISTANCE
(
    GNAME       VARCHAR(3)
,   CNAME       VARCHAR(3)
,   DIST        NUMERIC(5,3)
)

DROP TABLE IF EXISTS #RESULT
CREATE TABLE #RESULT
(
    GNAME       VARCHAR(3)
,   CNAME       VARCHAR(3)
,   DIST        NUMERIC(5,3)
--, FINRNK      INT
)
--INSERT INTO #DISTANCE
--VALUES 
--  ('E1', 'C1', 1  )       
--, ('E1', 'C2', 2  )
--, ('E1', 'C3', 3  )
--, ('E1', 'C4', 5  )
--, ('E2', 'C1', 2.5)   
--, ('E2', 'C2', 4  )
--, ('E2', 'C3', 3.5)
--, ('E2', 'C4', 6  )
--, ('E3', 'C4', 7  )       
--, ('E3', 'C5', 6  )
--, ('E3', 'C6', 4  )
---- CONCLUSIONG; SCENARIO WORKS, VALUES SHIFT.


--INSERT INTO #DISTANCE
--VALUES 
--  ('E1', 'C1', 1  )       
--, ('E1', 'C2', 2  )
--, ('E1', 'C3', 3  )
--, ('E1', 'C4', 5  )
--, ('E2', 'C7', 2.5)   
--, ('E2', 'C8', 4  )
--, ('E2', 'C9', 3.5)
--, ('E2', 'C4', 6  )
--, ('E3', 'C4', 7  )       
--, ('E3', 'C5', 6  )
--, ('E3', 'C6', 4  )
----,   ('E3', 'C3', 1  )
-- CONCLUSIONG; SCENARIO WORKS, VALUES SHIF

INSERT INTO #DISTANCE
VALUES 
    ('E1', 'C1', 1  )       
,   ('E1', 'C2', 2  )
,   ('E1', 'C3', 3  )
,   ('E1', 'C4', 4  ) -- switch with E2 dist to switch groups in results
,   ('E2', 'C1', 2.5)   
,   ('E2', 'C2', 4  )
,   ('E2', 'C3', 3.5)
,   ('E2', 'C4', 6  ) -- switch dist with e1 to move groups in results
,   ('E3', 'C4', 7  )       
,   ('E3', 'C5', 6  )
,   ('E3', 'C6', 4  )
,   ('E3', 'C3', 1  )

--INSERT INTO #DISTANCE
--VALUES 
--  ('E1', 'C1', 1  )       
--, ('E1', 'C2', 2  )
--, ('E1', 'C3', 3  )
--, ('E1', 'C4', 5  )
--, ('E2', 'C7', 2.5)   -- CHANGE CVAL TO UNIQUE TO PULL THROUGH    
--, ('E2', 'C8', 4  )   -- CHANGE CVAL TO UNIQUE TO PULL THROUGH
--, ('E2', 'C11', 3.5)  -- CHANGE CVAL TO UNIQUE TO PULL THROUGH
--, ('E2', 'C4', 6  )
--, ('E3', 'C9', 3  ) -- NEW
--, ('E3', 'C4', 7  )       
--, ('E3', 'C5', 6  )
--, ('E3', 'C6', 4  )
-- SCENARIO CORRECT. C4 SHOULD NOT BE PULLED IN.
GO

BEGIN
    ;
    WITH 
     NOROOM AS (
        SELECT GNAME FROM #RESULT GROUP BY GNAME HAVING COUNT(1) >= 3 -- HAS TO BE NOROOM BECAUSE 'HASROOM' WOULD not make sense in first pass.
    )
    ,X AS (
        SELECT 
            A.CNAME
        ,   A.DIST
        ,   A.GNAME
        ,   RNK = RANK()        OVER (PARTITION BY A.GNAME ORDER BY A.DIST ASC)
        ,   RNK2 = ROW_NUMBER() OVER (ORDER BY A.DIST ASC)
        ,   #RESULT.CNAME AS RSLT
        FROM #DISTANCE A
        LEFT JOIN #RESULT ON A.CNAME = #RESULT.CNAME
        LEFT JOIN NOROOM ON A.GNAME = NOROOM.GNAME
        WHERE NOROOM.GNAME IS NULL AND #RESULT.CNAME IS NULL --ORDER BY RNK
        )
        --SELECT * FROM X
    , Y AS (
        SELECT X.CNAME, MINRNK2 = MIN(X.RNK2) 
        FROM X 
        --LEFT JOIN #RESULT B ON X.CNAME = B.CNAME
        GROUP BY  X.CNAME
    )
    --  select * from y
    --  SELECT * FROM x JOIN Y ON X.CNAME = Y.CNAME AND X.RNK2 = Y.MINRNK2
    INSERT INTO #RESULT
    SELECT X.GNAME, X.CNAME, X.DIST FROM X 
     JOIN Y AS SQ ON X.CNAME = SQ.CNAME AND X.RNK2 = SQ.MINRNK2
    WHERE X.RNK <=3 --AND   #RESULT.GNAME IS NULL

     PRINT 'LOOP'
END
GO 10
     SELECT * FROM #RESULT ORDER BY GNAME

Answer 2

您似乎想要每个cname 中dist 最少的行。很简单：

select gname, cname, dist
from
(
  select
    gname, cname, dist, rank() over (partition by cname order by dist) as rnk
  from mytable
)as r
where rnk = 1;

我在这里使用RANK 来处理关系。因此，当cname 有两行具有相同的最小值dist 时，您将获得这两行的cname。如果您希望每个cname 只允许一行，则必须使用ROW_NUMBER 而不是RANK，但您还必须决定要显示哪一个gname。