alter proc zhilincafen @itemno varchar(50),@zilinno varchar(50),@cafenshu int
as
delete from  zilintab_lc_d where itemno =@itemno

declare @sql varchar(8000)     set @sql=''
declare @sql2 varchar(8000)    set @sql2=''
declare @num int               
declare @xuhao varchar(10)     set @xuhao='001'

declare @i int  set @i=0  
while @i<=25
begin

    set @sql2='select @num=z'+cast(@i as varchar(10))+' from zilintab where itemno= '''+@zilinno+''''
select @sql2
    exec(@sql2)

    while @num>0
    begin            
        if @num>@cafenshu
        begin
        set @sql='insert into zilintab_lc_d(zilinno,z'+cast(@i as varchar(10))+') select '+@zilinno+'-'+right('00'+@xuhao,3)+','+cast(@num as varchar(10))        
        exec(@sql)
        end
        else if @num<@cafenshu
        begin 
        set @sql='insert into zilintab_lc_d(zilinno,z'+cast(@i as varchar(10))+') select '+@zilinno+'-'+right('00'+@xuhao,3)+','+cast(@cafenshu as varchar(10))        
        exec(@sql)
        end
    set @xuhao=@xuhao+1
    set @num=@num-@cafenshu        
    end

    set @i=@i+1
end

--#1.
DECLARE @tempTable TABLE
(
    num INT
)

--#2.
DELETE FROM @tempTable
INSERT INTO @tempTable
EXEC(sqlString)
--#3.
SELECT @num = num FROM @tempTable

------解决方案--------------------
lz 外面还用到 @num  那就这样  在懂太语句中声明一个变量 把@@num传进去
SQL code

set sql2='daclare @num int'
set sql2=sql2+char(10)+'set @num='+@num+''
set @sql2=sql2+char(10)+'select @num=z'+cast(@i as varchar(10))+' from zilintab where itemno= '''+@zilinno+''''

------解决方案--------------------
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lz 外面还用到 @num 那就这样 在懂太语句中声明一个变量 把@@num传进去

SQL code

set sql2='daclare @num int'
set sql2=sql2+char(10)+'set @num='+@num+''
set @sql2=sql2+char(10)+'select @num=z'+cast(@i as varchar(10))+' from……