一个sql

zuochc

用DINSTINCT可以更简单呢:

SQL> select name
  2     from test
  3     group by name
  4     having count(distinct case when val in (1, 3, 5) then val end) = 3;

NAME
----------
a
b

akershao

select name,count(*)
from table_test5
where number  in(1,3,5)
group by name
having count(*)=3;

sensilo

decode 函数 is so strong

小K47

使用dense_rank()函数,然后结果等于3就是了~