用什么方法实现呢?

Question

DECLARE @L INT,@NAME VARCHAR(10),@sql varchar(1000)
SET @L=100
SET @NAME='*20*4'
select @sql='select '+cast(@L as varchar(38))+@NAME
exec(@sql)-----------
8000(1 行受影响)

Answer 1

更正一下:
DECLARE @L INT,@NAME VARCHAR(10)
SET @L=100
SET @NAME='20*4'
SELECT @L*@NAME最后结果等于:8000

Answer 2

DECLARE @L INT,@NAME VARCHAR(100)
SET @L=100
SET @NAME='*20*4'
set @name='select '+ltrim(@l)+@nameselect @nameexec(@name)8000

Answer 3

再问小虫兄:
前题:   
有三个表(T_B1,T_B2,T_B3),数据如下:   
T_B1:   T_ID                   T_CONT   
L001                           50   
L002                           120   
L003                           75   T_B2(其中T_B2的T_ID与T_B1的T_ID是相关联的,存在一对多关系):   T_ID               N_ID               N_NAME   
L001                   1                           *0.05*2   
L001                   2                           *0.1*2   
L002                   1                           *0.14*3   
L002                   2                           *0.25*4   
L002                   3                           *0.35*4   
L003                   1                           *0.45*2   
L003                   2                           *0.6*2   T_B3(其中T_B3的N_ID字段与T_B2的N_ID是相关联的存在一对多关系):   
N_ID           NAME               M_COUNT   
1                           李四                   T_B2表中的N_NAME字段*T_B1表的T_CONT字段   
1                           王五   
3                           张六   
2                           陈七   要解决的问题:   
写一存储过程达到以下结果:   
N_ID  T_ID  NAME  M_COUNT   
1     L001  李四     5      --50*0.05*2               T_B2表中的N_NAME字段*T_B1表的T_CONT字段   
1     L002  王五    50.4           --120*0.14*3   
3     L002  张六    168               ---120*0.35*4   
2     L003  陈七    90                   ----75*0.6*2