CODEWITHSUNDEEP
sqloracle-databasehierarchical-data
user3588186
user3588186

Reputation: 3

How to get information of root node for leaf node using SQL

Assume I have a table that has the following structure:

---------------------------------------------------
TREE_GID | NODE_ID | PARENT_NODE_ID | TREE_NAME   |
---------------------------------------------------
1        | 1       |                | A           |
1        | 2       | 1              |             |
1        | 3       | 1              |             |
1        | 4       | 2              |             |
1        | 5       | 4              |             |
1        | 6       |                | B           |
1        | 7       | 6              |             |
2        | 1       |                | C           |
2        | 2       | 1              |             |
---------------------------------------------------

Note:

I would like to display TREE_NAME in each nodes. The output should be like:

---------------------------------
TREE_ID | NODE_ID | TREE_NAME   |
---------------------------------
1       | 1       | A           |
1       | 2       | A           |
1       | 3       | A           |
1       | 4       | A           |
1       | 5       | A           |
1       | 6       | B           |
1       | 7       | B           |
2       | 1       | C           |
2       | 2       | C           |
---------------------------------

How can I get them?

Upvotes: 0

Views: 1785

Answers (1)

Joachim Isaksson
Joachim Isaksson

Reputation: 180917

A recursive common table expression will do it in a fairly straight forward way;

WITH cte(tree_gid, node_id, tree_name) AS (
  SELECT tree_gid, node_id, tree_name FROM mytable WHERE tree_name IS NOT NULL
  UNION ALL
  SELECT m.tree_gid, m.node_id, cte.tree_name
  FROM mytable m 
  JOIN cte ON cte.node_id = m.parent_node_id AND cte.tree_gid = m.tree_gid 
)
SELECT * FROM cte
ORDER BY tree_gid, node_id;

EDIT: You could also use CONNECT BY as your commented link describes;

SELECT tree_gid, node_id, CONNECT_BY_ROOT tree_name
  FROM mytable
  START WITH tree_name IS NOT NULL
  CONNECT BY PRIOR node_id = parent_node_id AND PRIOR tree_gid = tree_gid
  ORDER SIBLINGS BY tree_gid, node_id;

An SQLfiddle to test both.

Upvotes: 1

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