我创建了一个简单的例子来说明PostgreSQL中使用递归查询的传递闭包。
但是,我的递归查询有问题。我还不熟悉语法,所以这个请求可能完全不适合我,为此我提前道歉。如果运行查询,您将看到节点1在路径结果中重复自身。有人能帮我弄清楚如何调整SQL吗?
/* 1
/ \
2 3
/ \ /
4 5 6
/
7
/ \
8 9
*/
create table account(
acct_id INT,
parent_id INT REFERENCES account(acct_id),
acct_name VARCHAR(100),
PRIMARY KEY(acct_id)
);
insert into account (acct_id, parent_id, acct_name) values (1,1,'account 1');
insert into account (acct_id, parent_id, acct_name) values (2,1,'account 2');
insert into account (acct_id, parent_id, acct_name) values (3,1,'account 3');
insert into account (acct_id, parent_id, acct_name) values (4,2,'account 4');
insert into account (acct_id, parent_id, acct_name) values (5,2,'account 5');
insert into account (acct_id, parent_id, acct_name) values (6,3,'account 6');
insert into account (acct_id, parent_id, acct_name) values (7,4,'account 7');
insert into account (acct_id, parent_id, acct_name) values (8,7,'account 8');
insert into account (acct_id, parent_id, acct_name) values (9,7,'account 9');
WITH RECURSIVE search_graph(acct_id, parent_id, depth, path, cycle) AS (
SELECT g.acct_id, g.parent_id, 1,
ARRAY[g.acct_id],
false
FROM account g
UNION ALL
SELECT g.acct_id, g.parent_id, sg.depth + 1,
path || g.acct_id,
g.acct_id = ANY(path)
FROM account g, search_graph sg
WHERE g.acct_id = sg.parent_id AND NOT cycle
)
SELECT path[1] as Child,parent_id as Parent,path || parent_id as path FROM search_graph
ORDER BY path[1],depth;
最佳答案
您可以在几个地方进行简化(假设acct_id
和parent_id
是NOT NULL
):
WITH RECURSIVE search_graph AS (
SELECT parent_id, ARRAY[acct_id] AS path
FROM account
UNION ALL
SELECT g.parent_id, sg.path || g.acct_id
FROM search_graph sg
JOIN account g ON g.acct_id = sg.parent_id
WHERE g.acct_id <> ALL(sg.path)
)
SELECT path[1] AS child
, path[array_upper(path,1)] AS parent
, path
FROM search_graph
ORDER BY path;
列
acct_id
,depth
,cycle
只是查询中的噪声。WHERE
条件必须提前一步退出递归,然后从顶部节点的重复条目出现在结果中。在你的原著里是“一个一个”的。剩下的是格式化。
如果您知道图表中唯一可能的圆是自参考,我们可以有更便宜的:
WITH RECURSIVE search_graph AS (
SELECT parent_id, ARRAY[acct_id] AS path, acct_id <> parent_id AS keep_going
FROM account
UNION ALL
SELECT g.parent_id, sg.path || g.acct_id, g.acct_id <> g.parent_id
FROM search_graph sg
JOIN account g ON g.acct_id = sg.parent_id
WHERE sg.keep_going
)
SELECT path[1] AS child
, path[array_upper(path,1)] AS parent
, path
FROM search_graph
ORDER BY path;
SQL Fiddle.
注意,对于带有修饰符(如
varchar(5)
)的数据类型(至少pg v9.4)会有问题,因为数组连接会丢失修饰符,但rCTE坚持类型完全匹配:Surprising results for data types with type modifier