一个关于IP查询的SQL语句,该怎么解决

一个关于IP查询的SQL语句

有这样的张表结构
我现在有一个IP:211.136.168.140
我想查询它属于哪个ISP,这个查询语名怎么写。。。

------解决方案--------------------
改结构吧,效率会提高很多的

file data
ip1 58.211.0.1 
ip2 58.211.1.0 
intip1 986906625 
intip2 986906880 
ISP 电信
.......

intip1=256*256*256*58+256*256*211+256*0+1
intip2=256*256*256*58+256*256*211+256*1+0

转化 211.136.168.140 为 int
intip=256*256*256*211+256*256*136+256*168+140

select ISP from table where intip>=intip1 and intip<=intip2
------解决方案--------------------
SQL code
create table iptable(ip_start_one varchar(3),ip_start_two varchar(3),ip_start_three varchar(3),ip_start_four varchar(3),
ip_end_one varchar(3),ip_end_two varchar(3),ip_end_three varchar(3),ip_end_four varchar(3),isp nvarchar(15))
insert into iptable select '211','136','150','67','211','136','172','105','移动GPRS'
go
declare @ip varchar(15)
set @ip='211.136.168.140'
declare @ip1 varchar(3),@ip2 varchar(3),@ip3 varchar(3),@ip4 varchar(3)
set @ip1=left(@ip,charindex('.',@ip)-1)
set @ip=right(@ip,len(@ip)-charindex('.',@ip))
set @ip2=left(@ip,charindex('.',@ip)-1)
set @ip=right(@ip,len(@ip)-charindex('.',@ip))
set @ip3=left(@ip,charindex('.',@ip)-1)
set @ip4=right(@ip,len(@ip)-charindex('.',@ip))
select isp from IPTABLE
where 1=
(case when @ip1 between ip_start_one and ip_end_one then
    (case when ip_start_one<ip_end_one then 1 else
        (case when @ip2 between ip_start_two and ip_end_two then
            (case when ip_start_two<ip_end_two then 1 else
                (case when @ip3 between ip_start_three and ip_end_three then
                    (case when ip_start_three<ip_end_three then 1 else
                        (case when @ip4 between ip_start_four and ip_end_four then 1 else 0 end)
                        end)
                    else 0 end)
                end)
            else 0 end)
        end)
    else 0 end)
/*
isp
---------------
移动GPRS

(1 行受影响)

*/
go
drop table iptable