如何通过返回所有层次结构级别的select语句中的分支来订购层次结构树？

Question

This is my current SQL:

SELECT filter_id, parent_id, level_id, match 
FROM t_filters
ORDER BY level_id;  

It shows the following:

ID      PARENT  RANK    FILTER VALUE
10004           1       HQ
10002           1       Finance
10006           1       IT
10003   10006   2       HQ - Central
10005   10004   2       HQ - Non Core & Legacy
10001   10005   3       Non Core

My first attempt here orders the records by the hierarchy_level_id, which is basically the rank.

Note the parent field is null where the record is at the top of the hierarchy.

Also, the branch can have any length from 1 to 8 entries (8 is the current max but could theoretically increase by 1 or 2).

What I want to show is the following, which should use the parent field to collate the individual branches of the hierarchy together:

ID      PARENT  RANK    FILTER VALUE
10004           1       HQ
10005   10004   2       HQ - Non Core & Legacy
10001   10005   3       Non Core
10002           1       Finance
10006           1       IT
10003   10006   2       HQ - Central

Answer 1

The following solution orders siblings by id . In your comments, you've mentioned wanting to order siblings by (filter) value . Just replace the relevant expression to achieve this.

Use recursive SQL, Oracle syntax:

SELECT *
FROM t_filters
START WITH parent IS NULL
CONNECT BY parent = PRIOR id
ORDER SIBLINGS BY id

Alternatively, SQL standard syntax (the standard and some databases would require a RECURSIVE keyword, but Oracle doesn't allow it). A bit more tedious, but more extensible:

WITH /* RECURSIVE */ r (id, parent, rank, value, path) AS (
  SELECT id, parent, rank, value, '' || id
  FROM t_filters
  WHERE parent IS NULL

  UNION ALL

  SELECT f.id, f.parent, f.rank, f.value, r.path || '/' || f.id
  FROM r
  JOIN t_filters f ON r.id = f.parent
)
SELECT *
FROM r
ORDER BY path