收藏的MSSQL求連續ID內數量合計-常見解決方案

代码

-- > Title  : MSSQL求連續ID內數量合計-常見解決方案

-- > Author : wufeng4552

-- > Date   : 2009-12-04

if   object_id ( ' tb ' ) is   not   null   drop   table tb

go

create   table tb(ID varchar ( 10 ),Num decimal ( 10 , 2 ))

insert tb

select   ' 001 ' , 200.00   union   all

select   ' 002 ' , 200.00   union   all

select   ' 003 ' , 300.00   union   all

select   ' 007 ' , 500.00   union   all

select   ' 008 ' , 800.00   union   all

select   ' 009 ' , 200.00   union   all

select   ' 012 ' , 100.00   union   all

select   ' 013 ' , 100.00   union   all

select   ' 014 ' , 200.00   union   all

select   ' 017 ' , 100.00   union   all

select   ' 018 ' , 400.00   union   all

select   ' 019 ' , 300.00

-- >方法 1 臨時表

if   object_id ( ' tempdb..#t1 ' ) is   not   null

drop   table #t1

if   object_id ( ' tempdb..#t2 ' ) is   not   null

drop   table #t2

go

-- 取出起號

select cnt = identity ( int , 1 , 1 ), *   into #t1 from tb t where   not   exists ( select   1   from tb where id = t.id - 1 ) order   by t.id

-- 取出止號

select cnt = identity ( int , 1 , 1 ), *   into #t2 from tb t where   not   exists ( select   1   from tb where id = t.id + 1 ) order   by t.id

select n. [ start ] + ' - ' + n. [ end ] 起止號, sum (num)合計

from tb m,

( select a.ID [ start ] ,b.ID [ end ]   from #t1 a,#t2 b where a.cnt = b.cnt) n

where m.ID between n. [ start ]   and n. [ end ]

group   by n. [ start ] + ' - ' + n. [ end ]

/*

起止號                  合計

--------------------- ---------------------------------------

001-003               700.00

007-009               1500.00

012-014               400.00

017-019               800.00

 

(4 個資料列受到影響)

 

*/

-- 方法 2

select   case   when   min (t.id) != max (t.id) then   min (t.id) + ' - ' + max (t.id) else   min (t.id) end 起止號,

       sum (num)合計

from (

select ID,

       cnt = cast (ID as   int ) - ( select   count ( * ) from tb n where m.ID > n.ID),

       num

from tb m

)t group   by cnt

/*

起止號                  合計

--------------------- ---------------------------------------

001-003               700.00

007-009               1500.00

012-014               400.00

017-019               800.00

 

(4 個資料列受到影響)

 

*/

-- 方法3

select   case   when   min (t.id) != max (t.id) then   min (t.id) + ' - ' + max (t.id) else   min (t.id) end 起止號,

       sum (num)合計

from (

select id,cnt = id - row_number() over ( order   by   getdate ()),num from tb

)t group   by cnt

/*

起止號                  合計

--------------------- ---------------------------------------

001-003               700.00

007-009               1500.00

012-014               400.00

017-019               800.00

 

(4 個資料列受到影響)

 

*/