考考你的SQL水平

select * from
(select * from test where hobby = 'basketball') as a join
(select * from test where hobby = 'badminton') as b
on a.name = b.name

select * from (
select name,
cont=sum(case  when hobby='basketball' or hobby='badminton' then 1 else 0 end)
from test  group by name
)t where t.cont>1  这才是正解

select NAME from (select * from test where hobby in ('basketball','badminton')) group by name having count(*)>1

select
distinct t1.*
from test t1
inner join test t2 on t1.name = t2.name and t2.hobby in ('basketball','badminton')
inner join test t3 on t2.name = t3.name and t3.hobby in ('basketball','badminton')
where t2.hobby !=t3.hobby

select t4.*
from test t4
where exists (select 1 from test t2 
inner join test t3 on t2.name = t3.name and t3.hobby in ('basketball','badminton')
where t2.hobby !=t3.hobby and t2.hobby in ('basketball','badminton') and t2.name=t4.name)

HAVING COUNT 扩展起来比较方便 

select * from test t
where
exists(select 'z' from test z where z.name=t.name and t.hobby<>z.hobby and z.hobby in('basketball','badminton'))
and t.hobby in('basketball','badminton')

select t1.* from test t1, test t2 where t1.name=t2.name
and(
t1.hobby='basketball' and t2.hobby='badminton' or
t2.hobby='basketball' and t1.hobby='badminton'
)order by t1.name, t1.hobby

如果第9行也想返回：

select t0.* from test t0, test t1, test t2
where t0.name=t1.name and t1.name=t2.name
and t1.hobby='basketball' and t2.hobby='badminton'
order by t0.name, t0.hobby

select name from test where HOBBY in ('basketball','badminton')
group by name having count(1)>1
前提条件是没重复数据

select name
from testbb a
where hobby = 'basketball'
and exists (
select 1
from testbb b
where a.name = b.name
and b.hobby = 'badminton'
)

=====================
倒，发了才发现跟楼上一哥们的一样，惭愧了。