with cte_name as (
SELECT 컬럼1, 컬럼2, ...
FROM 테이블명
WHERE 조건
)
select 컬럼1, 컬럼2, ...
from cte_name
with heavy_user as (
select id, name, host_id
from places
group by host_id
having count(host_id) >= 2
)
select p.id, p.name, p.host_id
from places p
join heavy_user hy on hy.host_id = p.host_id
order by p.id