Zak的答案很接近,但根文件夾沒有很好地彙總。以下工作:
with par_child as (
select 1 as parent_id, 2 as child_id, 1 as child_class
union all select 1, 3, 1
union all select 1, 4, 2
union all select 2, 5, 1
union all select 2, 6, 2
union all select 2, 10, 2
union all select 3, 11, 2
union all select 3, 7 , 2
union all select 5, 8 , 2
union all select 5, 9 , 2
union all select 5, 12, 1
union all select 5, 13, 1
)
, child_cnt as
(
select parent_id as root_parent_id, parent_id, child_id, child_class, 1 as lvl from par_child union all
select cc.root_parent_id, pc.parent_id, pc.child_id, pc.child_class, cc.lvl + 1 as lvl from
par_child pc join child_cnt cc on (pc.parent_id=cc.child_id)
),
distinct_folders as (
select distinct child_id as folder_id from par_child where child_class=1
)
select root_parent_id, count(child_id) as cnt from child_cnt where child_class=2 group by root_parent_id
union all
select folder_id, 0 from distinct_folders df where not exists (select 1 from par_child pc where df.folder_id=pc.parent_id)