树结构和递归

Question

使用PostgreSQL 8.4.14数据库,我有一个表格树结构的表,如下例所示:

CREATE TABLE unit (
    id bigint NOT NULL PRIMARY KEY,
    name varchar(64) NOT NULL,
    parent_id bigint,
    FOREIGN KEY (parent_id) REFERENCES unit (id)
);
INSERT INTO unit VALUES (1, 'parent', NULL), (2, 'child', 1)
                      , (3, 'grandchild A', 2), (4, 'grandchild B', 2);

 id |    name      | parent_id 
----+--------------+-----------
  1 | parent       |          
  2 | child        |         1
  3 | grandchild A |         2
  4 | grandchild B |         2

我想为这些单元创建一个访问控制列表,其中每个单元可能拥有自己的ACL,或者从具有自己的ACL的最近的祖先继承它.

CREATE TABLE acl (
    unit_id bigint NOT NULL PRIMARY KEY,
    FOREIGN KEY (unit_id) REFERENCES unit (id)
);
INSERT INTO acl VALUES (1), (4);

 unit_id 
---------
       1
       4

我正在使用一个视图来确定一个单元是否从祖先继承它的ACL:

CREATE VIEW inheriting_acl AS
    SELECT u.id AS unit_id, COUNT(a.*) = 0 AS inheriting
    FROM unit AS u
    LEFT JOIN acl AS a ON a.unit_id = u.id
    GROUP BY u.id;

 unit_id | inheriting 
---------+------------
       1 | f
       2 | t
       3 | t
       4 | f

我的问题是:我怎样才能得到最近的单元,它不是从祖先继承ACL？我的预期结果应类似于下表/视图:

 unit_id | acl 
---------+------------
       1 | 1
       2 | 1
       3 | 1
       4 | 4

Answer 1

使用递归CTE的查询可以完成这项工作.需要PostgreSQL 8.4或更高版本:

WITH RECURSIVE next_in_line AS (
    SELECT u.id AS unit_id, u.parent_id, a.unit_id AS acl
    FROM   unit u
    LEFT   JOIN acl a ON a.unit_id = u.id

    UNION  ALL
    SELECT n.unit_id, u.parent_id, a.unit_id
    FROM   next_in_line n
    JOIN   unit u ON u.id = n.parent_id AND n.acl IS NULL
    LEFT   JOIN acl a ON a.unit_id = u.id
    )
SELECT unit_id, acl
FROM   next_in_line
WHERE  acl IS NOT NULL
ORDER  BY unit_id

第二站的休息状态UNION是n.acl IS NULL.这样,一旦acl找到,查询就会停止遍历树.
在最后,SELECT我们只返回找到的行acl.瞧.

暂且不说:使用通用的非描述性id列名作为反模式.可悲的是,默认情况下,一些ORM会这样做.调用它unit_id,您不必一直在查询中使用别名.