Question

我创建了一个简单的例子，用PostgreSQL中的递归查询来说明传递闭包。

然而，我的递归查询出了点问题。我还不熟悉语法，所以这个请求可能完全是我的noobish，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。有人可以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;

Answer 1

您可以在多个地方进行简化（假设acct_id和parent_id为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只是您查询中的噪音。
WHERE条件必须提前一步退出递归，之前来自顶级节点的重复条目在结果中。这是你原版中的“一个一个”。

其余的是格式化。

如果您知道图表中唯一可能的圈子是自我参考，我们可以更便宜：

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.

请注意，对于带有修饰符的数据类型（如varchar(5)），会出现问题（至少高达pg v9.4），因为数组连接丢失了修饰符，但rCTE坚持完全匹配类型：

Surprising results for data types with type modifier

Answer 2

您已将帐户1设置为其自己的父级。如果您将该帐户的父级设置为null，则可以避免将该帐户设置为开始和结束节点（逻辑设置的方式将包括一个循环，但之后不会添加到该循环，似乎合情合理）。将最终的“路径”列更改为case when parent_id is not null then path || parent_id else path end之类的内容看起来也更好一些，以避免在最后使用null。

用于传递闭包的递归查询

2 个答案: