通過顯示葉節點的路徑來顯示樹的層次結構

Question

我想在oracle中編寫查詢，以便可以將我的樹層次結構排成一行，並且僅顯示最后一個節點路徑，就像我有一個ID及其父ID。

id                parent_id
1                 null
2                 1
3                 2
4                 3
5                 4
6                 5

並且輸出應該像路徑一樣成一行

1-2-3-4-5-6

Answer 1

使用函數sys_connect_by_path()和偽列connect_by_is_leaf ：

select id, ltrim(sys_connect_by_path(id, '-'), '-') path
  from test where connect_by_isleaf = 1
  connect by prior id = parent_id
  start with parent_id is null

輸出和SQLFiddle ：

ID   PATH
---  ---------------
  6  1-2-3-4-5-6

---

編輯： where子句在這里很自然，但是由於某些原因您絕對不想要它。 您可以使用：

select id, path from (
  select id, ltrim(sys_connect_by_path(id, '-'), '-') path, connect_by_isleaf leaf
    from test connect by prior id = parent_id
    start with parent_id is null)
  connect by 1=0 start with leaf = 1

Answer 2

select  LISTAGG(id, '-') WITHIN GROUP (ORDER BY id) from trea_sort connect by PRIOR  id = parent start with parent is null