Question

按T分组行，并在每个组中查找最大或最小的行（如果值为负），该行中其他行的总和，并删除该行（每个组一个），如果group没有足够的元素来查找sum或者足够但没有行指示其他的总和没有发生

CREATE TABLE Test (
    T   varchar(10),
    V   int
);
INSERT INTO Test
    VALUES ('A',  4), 
           ('B', -5), 
           ('C',  5), 
           ('A',  2), 
           ('B', -1), 
           ('C', 10), 
           ('A',  2), 
           ('B', -4), 
           ('C',  5), 
           ('D',  0);

预期结果：

Answer 1

与评论一样，要求似乎很奇怪。下面的代码假定求和已经预先填充，只要最高值不是0就删除最大值/最小值。

if object_id('tempdb..#test') is not null
drop table #test

CREATE TABLE #Test (
    T   varchar(10),
    V   int
);
INSERT INTO #Test
    VALUES ('A', 4), ('B', -5), ('C', 5), ('A', 2), ('B', -1), ('C', 10), ('A', 2), ('B', -4), ('C', 5), ('D', 0);


if object_id('tempdb..#test2') is not null
drop table #test2

SELECT
T,
V,
ABS(V) as absV
INTO #TEST2
FROM #TEST
SELECT * FROM #TEST2

if object_id('tempdb..#max') is not null
drop table #max

SELECT
T,
MAX(absV) AS MaxAbsV
INTO #Max
FROM #TEST2
GROUP BY T
HAVING MAX(AbsV) != 0



DELETE #TEST2
FROM #TEST2
INNER JOIN #MAX ON #TEST2.T = #MAX.T AND #TEST2.absV = #Max.MaxAbsV

SELECT * FROM #TEST2
ORDER BY T ASC

Answer 2

; with cte as
(
    select  T, V, 
        R = row_number() over (partition by T order by ABS(V) desc),
        C = count(*) over (partition by T)
    from    Test 
)
delete  c
from    cte c
    inner join
    (
        select  T, S = sum(V)
        from    cte
        where   R   <> 1
        group by T
    ) s on  c.T = s.T
where   c.C >= 3 
and c.R = 1
and c.V = s.S

Answer 3

使用ABS而不存在

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Answer 4

首先通过检查SUM(V)是否为正，确定行是正数还是负数。然后确定最小值或最大值是否等于其他行的SUM，通过从SUM(V)减去MIN(V)如果为负，或MAX(V)如果为正：

DELETE t 
FROM Test t
INNER JOIN (
    SELECT
        T, 
        SUM(V) - CASE WHEN SUM(V) >= 0 THEN MAX(V) ELSE MIN(V) END AS ToDelete
    FROM Test 
    GROUP BY T
    HAVING COUNT(*) >= 3
) a
    ON a.T = t.T
    AND a.ToDelete = t.V

ONLINE DEMO

Answer 5

您可以使用以下查询来获取所需的输出： -

select * into #t1 from test

select * from
(
select TT.T as T,TT.V as V
from  test TT
JOIN
(select T,max(abs(V)) as V from #t1 
group by T) P
on TT.T=P.T
where abs(TT.V) <> P.V

UNION ALL


select A.T as T,A.V as V from test A
JOIN(
select T,count(T) as Tcount from test
group by T
having count(T)=1) B on A.T=B.T
) X order by T

drop table #t1

Answer 6

您正在寻找每个组的值，该值是所有组的其他值的总和。例如。 4（2,2,4）或-5（-5，-4，-1）。

这通常每组只有一条记录。但它可以是同一数字的多倍。以下是关系的例子：（0,0）或（-2,2,4,4），或（-2，-2,4,4,4）或（-10,3,3,3,3， 4）。

如您所见，您正在以任何方式查找等于该组总和的一半的值。（当然。我们正在寻找n + n，其中一个n在一个记录中，另一个n是所有其他记录的总和。）

唯一的特例是当组中只有一个值为零时。我们当然不想删除。

这是一个无法处理关系的更新语句，但会删除所有最大值而不只是一个：

delete from test
where 2 * v =
(
  select case when count(*) = 1 then null else sum(v) end
  from test fullgroup
  where fullgroup.t = test.t
);

为了处理关系，你需要人工行号，以便只删除所有候选人的一条记录。

with candidates as
(
  select t, v, row_number() over (partition by t order by t) as rn
  from
  (
    select 
      t, v, 
      sum(v) over (partition by t) as sumv, 
      count(*) over (partition by t) as cnt
    from test
  ) comparables
  where sumv = 2 * v and cnt > 1
) 
delete
from candidates
where rn = 1;

SQL小提琴：http://sqlfiddle.com/#!6/6d97e/1

Answer 7

查看以下查询是否有帮助：

DELETE [Audit].[dbo].[Test] FROM [Audit].[dbo].[Test] as AA
INNER JOIN (select T, 
      CASE 
          WHEN MAX(V) < 0 THEN MIN(V) 
          WHEN MIN(V) > 0 THEN MAX(V) ELSE MAX(V)
      END as MAX_V,
      CASE 
          WHEN SUM(V) > 0 THEN  SUM(V) - MAX(V)
          WHEN SUM(V) < 0 THEN  SUM(V) - MIN(V) ELSE SUM(V)
      END as SUM_V_REST
      from [Audit].[dbo].[Test] 
      Group by T 
      Having Count(V) > 1) as BB ON AA.T = BB.T and AA.V = BB.MAX_V

删除表中的行，这些行是每个组的其他行的总和

7 个答案: