SQL Query: Fetch ordered rows from a table

二次信任 提交于 2020-01-17 03:05:13

问题


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 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

回答1:


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;



回答2:


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


来源:https://stackoverflow.com/questions/6040933/sql-query-fetch-ordered-rows-from-a-table

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