SQL Query: Fetch ordered rows from a table

Following are some entries from a table:

id      r_id        a_id        p_id


1       9           9           0
2       9           105         108
3       9           102         9
4       9           106         105
开发者_运维百科5       9           108         102

Is it possible to get the following output using SQL query

1       9           9           0
3       9           102         9
5       9           108         102
2       9           105         108
4       9           106         105

The idea is to sort the rows in such a way that a row with p_id = x should come below the row with a_id = x.

I hope question makes sense.

Regards,

Mayank

EDIT:

I'm looking this for PostgreSql

• The root item has a p_id = 0
• There are no missing links

---

Use a recursive query (PostgreSQL version 8.4 or later):

/* test data:
CREATE TABLE foo (id, r_id, a_id, p_id) AS
    SELECT  1,9,9,0
    UNION ALL SELECT 2,9,105,108
    UNION ALL SELECT 3,9,102,9
    UNION ALL SELECT 4,9,106,105
    UNION ALL SELECT 5,9,108,102        
;
*/

-- the query you need:
WITH RECURSIVE sub(s_id, s_r_id, s_a_id, s_p_id, row) AS (
    SELECT id, r_id, a_id, p_id, 1 AS row FROM foo WHERE p_id = 0
UNION ALL
    SELECT id, r_id, a_id, p_id, (row + 1)  FROM foo JOIN sub ON s_a_id = p_id
)
SELECT * FROM sub ORDER BY row;

---

Following is adapted from a working SQL Server 2005 solution.

I have made some assumptions

• The root item has a p_id = 0
• There are no missing links

SQL Statement

;WITH RECURSIVE q AS (
    SELECT  *
            , 1 AS Level
    FROM    ATable 
    WHERE   p_id = 0
    UNION ALL
    SELECT  t.*
            , Level = Level + 1
    FROM    q
            INNER JOIN ATable t ON t.p_id = q.a_id          
)
SELECT  *
FROM    q
ORDER BY
        Level