您可以使用以下方法來獲取父母/根目錄的列表:
select parent_id as parent
from tbl x
where not exists (select 1 from tbl y where y.child_id = x.parent_id)
| PARENT |
|--------|
| 15 |
| 21 |
| 17 |
| 30 |
http://sqlfiddle.com/#!4/62c37/18/0
然後對於一個給定根,你可以運行以下命令:
select 21 as parent,
listagg(child_id, ' >>> ') within group(order by level) as children
from tbl
start with parent_id = 21
connect by prior child_id = parent_id
| PARENT | CHILDREN |
|--------|------------------------|
| 21 | 20 >>> 4 >>> 23 >>> 39 |
http://sqlfiddle.com/#!4/62c37/35/0
或者如果你不想帶孩子仁到一排,你可以得到類似的輸出到你給運行什麼:
select 21 as parent,
lpad(child_id,level*level,' ') as child
from tbl
start with parent_id = 21
connect by prior child_id = parent_id
| PARENT | CHILD |
|--------|-----------|
| 21 | 2 |
| 21 | 4 |
| 21 | 23 |
| 21 | 39 |
http://sqlfiddle.com/#!4/62c37/34/0
我不知道在一個查詢運行它的所有根源,但想我會至少給你等待另一個答案。
但是,如果總是隻有少數幾個孩子(即在10歲以下),那麼這應該是可行的,而不依賴於CONNECT BY,而是使用子查詢。
由於這一做法的一個例子,下面那張8級深,只要有從未超過8個孩子(你可以添加更多的潛艇,如果不是這種情況)在您的整個表應該工作:
with lvl1 as
(select x.parent_id, x.child_id
from tbl x
where not exists (select 1 from tbl y where y.child_id = x.parent_id)),
lvl2 as
(select x.parent_id, x.child_id
from tbl x
left join lvl1
on x.parent_id = lvl1.child_id),
lvl3 as
(select x.parent_id, x.child_id
from tbl x
left join lvl2
on x.parent_id = lvl2.child_id),
lvl4 as
(select x.parent_id, x.child_id
from tbl x
left join lvl3
on x.parent_id = lvl3.child_id),
lvl5 as
(select x.parent_id, x.child_id
from tbl x
left join lvl4
on x.parent_id = lvl4.child_id),
lvl6 as
(select x.parent_id, x.child_id
from tbl x
left join lvl5
on x.parent_id = lvl5.child_id),
lvl7 as
(select x.parent_id, x.child_id
from tbl x
left join lvl6
on x.parent_id = lvl6.child_id),
lvl8 as
(select x.parent_id, x.child_id
from tbl x
left join lvl7
on x.parent_id = lvl7.child_id)
select parent_id,
case when lag(child2,1) over (partition by parent_id order by parent_id) = child2 then null else child2 end as child2,
case when lag(child3,1) over (partition by parent_id order by parent_id) = child3 then null else child3 end as child3,
case when lag(child4,1) over (partition by parent_id order by parent_id) = child4 then null else child4 end as child4,
case when lag(child5,1) over (partition by parent_id order by parent_id) = child5 then null else child5 end as child5,
case when lag(child6,1) over (partition by parent_id order by parent_id) = child6 then null else child6 end as child6,
case when lag(child7,1) over (partition by parent_id order by parent_id) = child7 then null else child7 end as child7,
case when lag(child8,1) over (partition by parent_id order by parent_id) = child8 then null else child8 end as child8
from(
select distinct
lvl1.parent_id,
lvl2.parent_id as child2,
nvl(lvl3.parent_id,lvl2.child_id) as child3,
nvl(lvl4.parent_id,lvl3.child_id) as child4,
nvl(lvl5.parent_id,lvl4.child_id) as child5,
nvl(lvl6.parent_id,lvl5.child_id) as child6,
nvl(lvl7.parent_id,lvl6.child_id) as child7,
nvl(lvl8.parent_id,lvl7.child_id) as child8
from lvl1
left join lvl2
on lvl1.child_id = lvl2.parent_id
left join lvl3
on lvl2.child_id = lvl3.parent_id
left join lvl4
on lvl3.child_id = lvl4.parent_id
left join lvl5
on lvl4.child_id = lvl5.parent_id
left join lvl6
on lvl5.child_id = lvl6.parent_id
left join lvl7
on lvl6.child_id = lvl7.parent_id
left join lvl8
on lvl7.child_id = lvl8.parent_id
order by parent_id)
http://sqlfiddle.com/#!4/62c37/50/0
| PARENT_ID | CHILD2 | CHILD3 | CHILD4 | CHILD5 | CHILD6 | CHILD7 | CHILD8 |
|-----------|--------|--------|--------|--------|--------|--------|--------|
| 15 | 12 | 5 | 23 | (null) | (null) | (null) | (null) |
| 17 | 12 | 5 | 23 | (null) | (null) | (null) | (null) |
| 21 | 20 | 4 | 23 | (null) | (null) | (null) | (null) |
| 21 | (null) | (null) | 39 | (null) | (null) | (null) | (null) |
| 30 | (null) | (null) | (null) | (null) | (null) | (null) | (null) |
您正在使用什麼版本的Oracle? –
Gordon:Oracle 10g – Mojoy
。 。太糟糕了。 Oracle 11g具有遞歸CTE,我發現它比'connect by'更容易使用。 –