SQL查询结果为二维表
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1 ---测试数据--- 2 if object_id(‘[tb]‘) is not null drop table [tb] 3 go 4 create table [tb]([code] varchar(6),[Month] int,[Num] int) 5 insert [tb] 6 select ‘C00001‘,200401,3 union all 7 select ‘C00001‘,200402,1 union all 8 select ‘C00001‘,200403,1 union all 9 select ‘C00001‘,200404,3 union all 10 select ‘C00001‘,200405,3 union all 11 select ‘C00001‘,200604,1 union all 12 select ‘C00002‘,200401,3 union all 13 select ‘C00002‘,200402,2 union all 14 select ‘C00002‘,200404,1 union all 15 select ‘C00002‘,200405,1 union all 16 select ‘C9999‘,200401,5 union all 17 select ‘C9999‘,200402,2 union all 18 select ‘C9999‘,200403,2 19 go 20 21 ---查询--- 22 declare @sql varchar(8000) 23 select 24 @sql=isnull(@sql+‘,‘,‘‘) 25 +‘sum(case when [month]=‘+ltrim([month])+‘ then num else 0 end) as [‘+ltrim([month])+‘]‘ 26 from 27 (select distinct [month] from tb) t 28 29 exec (‘select code,‘+@sql+‘ from tb group by code‘) 30 31 ---结果--- 32 code 200401 200402 200403 200404 200405 200604 33 ------ ----------- ----------- ----------- ----------- ----------- ----------- 34 C00001 3 1 1 3 3 1 35 C00002 3 2 0 1 1 0 36 C9999 5 2 2 0 0 0 37 38 (3 行受影响)
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