急帮忙写SQL话语_Sql Server

急！！帮忙写SQL语句
我想取出分组后满足每条分组记录的第一条记录
如有1个字段

通讯地址
192.168.1.111,800
192.168.1.111,801
192.168.1.250,801
192.168.1.250,802
192.168.1.250,803
192.168.1.251,805

我现在想按前半部分分组然后取出每组分组后的第一条记录

既得到如下记录
通讯地址
192.168.1.111,800
192.168.1.250,801
192.168.1.251,805

有人能帮实现一下吗？写不出来呢现在只能分组
select substring(通讯地址,1,13) from A group by substring(通讯地址,1,13)

------解决方案--------------------

SQL code

--tryselect substring(通讯地址,1,13),min(substring(通讯地址,15,3)) from A group by substring(通讯地址,1,13)
------解决方案--------------------
SQL codedeclare @t table([通讯地址] varchar(17))insert @tselect '192.168.1.111,800' union allselect '192.168.1.111,801' union allselect '192.168.1.250,801' union allselect '192.168.1.250,802' union allselect '192.168.1.250,803' union allselect '192.168.1.251,805'select * from @t twhere right([通讯地址],3)=(select min(right([通讯地址],3)) from @t where left([通讯地址],13)=left(t.[通讯地址],13))/*通讯地址-----------------192.168.1.111,800192.168.1.250,801192.168.1.251,805*/--也可以用逗号做标记分割
------解决方案--------------------
SQL codeIF EXISTS (SELECT 1 FROM SYSOBJECTS WHERE name = 'tba')BEGIN    DROP TABLE tbaENDGOCREATE TABLE tba(    ID INT IDENTITY(1,1),    IP VARCHAR(100))GOINSERT INTO tbaSELECT '192.168.1.111,800' UNIONSELECT '192.168.1.111,801'UNIONSELECT '192.168.1.250,801'UNIONSELECT '192.168.1.250,802'UNIONSELECT '192.168.1.250,803'UNIONSELECT '192.168.1.251,805'GOSELECT * FROM tba AS A WHERE IP = (SELECT TOP 1 IP FROM tba WHERE LEFT(IP,CHARINDEX(',',IP)) = LEFT(A.IP,CHARINDEX(',',A.IP)))ID    IP1    192.168.1.111,8003    192.168.1.250,8016    192.168.1.251,805