select * from tablename where left(no,12) = '110309030301'   

select * from t
inner join
(select distinct left(患者编号,12) as Left12,count(*) as NUM from t group by left(患者编号,12)) a
on a.Left12=left(t.患者编号,12) where NUM >3

select *
from tablename
where 患者编号 like '110311027301%'

试试！

select 患者编号 from tablename where left(rtrim(ltrim(患者编号)),12)='110311027301'

select * form 表 where 患都编号 like '11031102730%'

这句话的意思就是匹配前是这个11031102730的,而后面那个%并没有限定位数,
如果你的编号后面是要限定位数的,有几位则用几个_下划线来代替,通配符

问题解决了谢谢大家


select * from musterroll where left(医疗编号,12) in
(
select 成员 from
(
select count(*) as 相同成员,left(医疗编号,12) as 成员 from musterroll
group by left(医疗编号,12)
) as t
where 相同成员 >1

select a.患者编号,a.姓名,a.性别 from 表名 a,
(select left(患者编号,12) from 表名 group by left(患者编号,12)) b
where left(a.患者编号,12)=b.患者编号 order by a.患者编号

select * from table1 where left(患者编号,12) in
(select left(患者编号,12) from table1 group by left(患者编号,12) having count(*) > 1) order by left(患者编号,12)

这样试试

刚换工作单位 没装SQL呢还 所以没有试试看是不是能成功执行 抱歉了