这个选择如何写

[yyb2000]

表如下
field1 autoid
10,15 1
2,3,9,151 2
7,8,1 3

已确定field1中的数字只出现1 次，现要求如下，如何写语句得到'15'所在记录的autoid，用like会出现'151'

[pengdali]

select * from ','+field+',' like ',15,'

[pengdali]

select * from 表 where ','+field+',' like ',15,'


或：

select * from 表 where charindex(',15,',','+field+',')>0

[lynx1111]

select * from tab where ','+field+',' like '%,15,%'

[zarge]

select autoid from 表 where charindex(',15,',','+field1+',')>0

[pengdali]

select * from 表 where ','+field+',' like ',15,'
更正：
select * from 表 where ','+field+',' like '%,15,%'

[aierong]

create table aiii(field1 varchar(100), autoid int)
insert into aiii select '10,15', 1
insert into aiii select '2,3,9,151', 2
insert into aiii select '7,8,1', 3


select autoid
from aiii
where field1 like '%,15%'
and substring(field1,patindex('%15%',field1)+2,1) not like '[0-9]'

[伍子V5]

select * from tablename where charindex(',15,',','+field1+',')>0

[aierong]

select * from aiii where charindex(',15,',','+field1+',')<>0