Question

我想在oracle中编写一个查询，以便我可以将我的树层次结构放在一行中，并且只显示最后一个节点路径，就像我有一个id及其父ID一样。

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

并且输出应该在一行中路径

1-2-3-4-5-6

Answer 1

使用函数sys_connect_by_path()和伪列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

输出和SQLFiddle：

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

编辑： where条款在这里很自然，但出于某种原因，您绝对不想要它。你可以使用：

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

通过显示叶节点的路径来显示树的层次结构

2 个答案: