Question

这是可能的，我有如下结果，想要填充或组合每个id

Num    id   RMA
item1   1   0
item1   1   0
item1   1   0
item1   8   1
item1   8   1
item1   8   1
item1   1   0
item1   1   0
item1   1   0
item2   8   1
item2   8   1
item2   8   1
item2   8   1
item2   8   1
item2   8   1
item2   8   1
item2   1   0

我想得到如下结果

item1   id1=3,id8=3,id1=3
item2   id8=7,id1=1

Answer 1

如果是SQL服务器：

您可以使用GROUP BY子句：

SELECT    Num
          ,id
          ,COUNT(RMA) AS Count_RMA
FROM      Table
GROUP BY  Num
          ,id

它会输出一个表格，该表格会计算每个唯一RMA和Num组合的每个id。

像这样：

Num    id   Count_RMA
item1   1   6
item1   8   3
item2   8   7
item2   1   1

Answer 2

它看起来像这样：

select t.*,
       stuff( (select ',' + cast(id as varchar(max)) + '=' + cast(cnt as varchar(max))
               from t tt
               where t.num = tt.num
               group by id
               for xml path ('')
              ), 1, 1, ''
             ) as 
from (select num from t) t

Answer 3

但是，您需要创建一些额外的步骤并不困难，请参阅下面的代码：

DECLARE @items TABLE (Num varchar(10) , id INT, RMA int , _group int default(0) )
declare @gIndcator int = 0 , @old_id int = 0 

insert @items (num,id,rma) select 'item1' ,    1 , 0
insert @items (num,id,rma) select 'item1' ,    1 , 0
insert @items (num,id,rma) select 'item1' ,    1 , 0
insert @items (num,id,rma) select 'item1' ,    8 , 1
insert @items (num,id,rma) select 'item1' ,    8 , 1
insert @items (num,id,rma) select 'item1' ,    8 , 1
insert @items (num,id,rma) select 'item1' ,    1 , 0
insert @items (num,id,rma) select 'item1' ,    1 , 0
insert @items (num,id,rma) select 'item1' ,    1 , 0
insert @items (num,id,rma) select 'item2' ,    8 , 1
insert @items (num,id,rma) select 'item2' ,    8 , 1
insert @items (num,id,rma) select 'item2' ,    8 , 1
insert @items (num,id,rma) select 'item2' ,    8 , 1
insert @items (num,id,rma) select 'item2' ,    8 , 1
insert @items (num,id,rma) select 'item2' ,    8 , 1
insert @items (num,id,rma) select 'item2' ,    8 , 1
insert @items (num,id,rma) select 'item2' ,    1 , 0


UPDATE @items
SET @gIndcator = _group = case when @old_id <> id then @gIndcator +  1 else @gIndcator end 
    , @old_id = id 


select * from @items

select distinct
    num
  , ids = stuff((
      select ',id'+convert(varchar(12),b.id) + '=' +convert(varchar(12),b._count)
      from (
SELECT num, id , _group, COUNT(1) _count
FROM  @items c
where a.Num = c.Num
group by Num, id,_group)  b
      where a.Num = b.Num
      order by b.Num,_group
      for xml path (''), type).value('.','nvarchar(max)')
      ,1,1,'')
from @items a

item1 id1 = 3，id8 = 3，id1 = 3

item2 id8 = 7，id1 = 1

Answer 4

;WITH CTE(Num , id, RMA)
AS
(
SELECT 'item1',   1 ,  0 UNION ALL
SELECT 'item1',   1 ,  0 UNION ALL
SELECT 'item1',   1 ,  0 UNION ALL
SELECT 'item1',   8 ,  1 UNION ALL
SELECT 'item1',   8 ,  1 UNION ALL
SELECT 'item1',   8 ,  1 UNION ALL
SELECT 'item1',   1 ,  0 UNION ALL
SELECT 'item1',   1 ,  0 UNION ALL
SELECT 'item1',   1 ,  0 UNION ALL
SELECT 'item2',   8 ,  1 UNION ALL
SELECT 'item2',   8 , 1  UNION ALL
SELECT 'item2',   8 ,  1 UNION ALL
SELECT 'item2',   8 ,  1 UNION ALL
SELECT 'item2',   8 ,  1 UNION ALL
SELECT 'item2',   8 ,  1 UNION ALL
SELECT 'item2',   8 ,  1 UNION ALL
SELECT 'item2',   1 ,  0
)
,Final
AS
(
SELECT Num,idval,cnt AS COUNT_RMA From
(
SELECT NUM,'    id'+CASt(id AS Varchar(2))AS idval,cnt, ROW_NUMBER()OVEr(Partition by NUm,Cnt order by Num)AS Seq
From
(
SELECT NUm,id,COUNT(Id)Over(Partition by Num ,id order by Num)Cnt from CTE
)Dt
)Fnl
where Fnl.Seq=1
)
SELECT DISTINCT num, STUFF((SELECT DISTINCT ', '+CAST(''+CAST(idval AS VARCHAR)+'='+CAST(COUNT_RMA AS VARCHAR) AS VARCHAR(100))
From Final I WHERE O.Num=I.Num foR XML PATH('')),1,1,'') As Requiredvalue
from Final O

计算项目并填充结果

4 个答案: