Question

我有一个如下表：

create table path(
  id integer,
  start varchar(255),
  finish varchar(255),
  client integer

);

INSERT INTO path(id, start, finish, client) VALUES (1, '1', '2', 1);
INSERT INTO path(id, start, finish, client) VALUES (2, '2', '3', 1);
INSERT INTO path(id, start, finish, client) VALUES (3, '3', '4', 1);
INSERT INTO path(id, start, finish, client) VALUES (4, '5', '6', 1);

我尝试了以下查询：

select string_agg(start || ' -> ' ||finish, ', ' order by id) from path group by client;

结果是这样的：

1 -> 2, 2 -> 3, 3 -> 4, 5 -> 6

如果两个序列值相同，如何获得以下结果：

1 -> 2 -> 3 -> 4, 5 -> 6

这里是Sql Fiddle link。

Answer 1

如我的评论中所述，递归CTE将产生此结果。 As an example：

WITH RECURSIVE reccte AS
(
  /*Recursive Seed - Records that will start recursion*/
  SELECT id, start, finish, client, start || '->' || finish as path, 1 as depth
  FROM path
  WHERE id = 1
  UNION ALL
  /*Recursive Term - This will be the statement that iterates until the join fails*/
  SELECT
    path.id, path.start, path.finish, path.client, reccte.path || '->' || path.finish, depth + 1
  FROM reccte 
    INNER JOIN path ON reccte.finish = path.start
  WHERE depth < 20 /*make sure we don't loop endlessly*/
)
SELECT * FROM reccte;


+----+-------+--------+--------+---------------+-------+
| id | start | finish | client |     path      | depth |
+----+-------+--------+--------+---------------+-------+
|  1 |     1 |      2 |      1 | 1->2          |     1 |
|  2 |     2 |      3 |      1 | 1->2->3       |     2 |
|  3 |     3 |      4 |      1 | 1->2->3->4    |     3 |
|  4 |     4 |      5 |      1 | 1->2->3->4->5 |     4 |
+----+-------+--------+--------+---------------+-------+

在此示例中，最后一个SELECT可以更改为SELECT path FROM reccte ORDER BY depth DESC LIMIT 1;，以替换您想要的记录。

已更新：

这里的解决方案仍然相同，我们只是调整了递归CTE和最终选择，以获取层次结构的不同完整分支/路径：

http://sqlfiddle.com/#!17/2892f/6

WITH RECURSIVE reccte AS
(
  SELECT id as origin, id, start, finish, client, start || ' -> ' || finish as path, 1 as depth
  FROM path
  WHERE start NOT IN (SELECT distinct finish FROM path)
  UNION ALL
  SELECT
    reccte.origin, path.id, path.start, path.finish, path.client, reccte.path || ' -> ' || path.finish, depth + 1
  FROM reccte 
    INNER JOIN path ON reccte.finish = path.start
  WHERE depth < 20 /*make sure we don't loop endlessly*/
)
SELECT * FROM (SELECT origin, path, depth, max(depth) OVER (PARTITION BY origin) as maxdepth FROM reccte) sub WHERE sub.maxdepth = depth;


+--------+------------------+-------+----------+
| origin |       path       | depth | maxdepth |
+--------+------------------+-------+----------+
|      1 | 1 -> 2 -> 3 -> 4 |     3 |        3 |
|      4 | 5 -> 6           |     1 |        1 |
+--------+------------------+-------+----------+

您也可以在该最终选择中在此处string_agg（），以获取更新问题中的确切输出：

http://sqlfiddle.com/#!17/2892f/5

SELECT STRING_AGG(path, ', ' order by origin) FROM (SELECT origin, path, depth, max(depth) OVER (PARTITION BY origin) as maxdepth FROM reccte) sub WHERE sub.maxdepth = depth;

+--------------------------+
|        string_agg        |
+--------------------------+
| 1 -> 2 -> 3 -> 4, 5 -> 6 |
+--------------------------+

如何在string_agg内合并序列重复的值

1 个答案: