Question

树的深度无限。例如：

+----+-------+--------------------+
| id | name  | parent_id |  value |
+----+-------+--------------------+
|  1 | test1 |           |    0   |
|  2 | test2 |     1     |    0   |
|  3 | test3 |     2     |    5   |
|  4 | test4 |     1     |    0   |
|  5 | test5 |     4     |    5   |
|  6 | test6 |     4     |    0   |
|  7 | test7 |     6     |    10  |
+----+-------+--------------------+

我想得到一个孩子的总价值。就像这样：

+----+-------+--------------------+
| id | name  | parent_id |  value |
+----+-------+--------------------+
|  1 | test1 |           |    20  |  =  test2.value + test4.value
|  2 | test2 |     1     |    5   |  =  test3.value
|  3 | test3 |     2     |    5   |  
|  4 | test4 |     1     |    15  |  =  test5.value + test6.value
|  5 | test5 |     4     |    5   |
|  6 | test6 |     4     |    10  |  =  test7.value
|  7 | test7 |     6     |    10  |
+----+-------+--------------------+

有什么建议吗？谢谢！

Answer 1

这是一个递归查询，希望能够解决您的问题：

WITH RECURSIVE tree (id, parent_id, cnt) AS (
    -- start from bottom-level entries
    SELECT id, parent_id, 0::bigint AS cnt
    FROM tbl t
    WHERE NOT EXISTS (
        SELECT id
        FROM tbl
        WHERE parent_id = t.id
    )

    UNION ALL

    -- join the next level, add the number of children to that of already counted ones
    SELECT t.id, t.parent_id, tree.cnt + (
            SELECT count(id) 
            FROM tbl 
            WHERE parent_id = t.id
        )
    FROM tbl t JOIN tree ON t.id = tree.parent_id
)
SELECT tree.id, max(tree.cnt) AS number_of_children
FROM tree 
-- use the JOIN if you need additional columns from tbl
-- JOIN tbl ON tree.id = tbl.id 
-- use the WHERE clause if you want to see root nodes only
-- WHERE parent_id IS NULL
GROUP BY tree.id
ORDER BY tree.id
;

我也制作了SQLFiddle。

Answer 2

见F.35.1.5。 connectby函数示例：

http://www.postgresql.org/docs/current/static/tablefunc.html

postgresql统计孩子的数量

2 个答案: