PostgreSQL合并递归查询和JOIN
PostgreSQL merge recursive query and JOIN
我有以下架构:
CREATE TABLE tbl_employee_team
(
employee_id int,
teams_id int
);
INSERT INTO tbl_employee_team
VALUES
(1, 2),
(1, 3),
(1, 4);
CREATE TABLE tbl_team_list_serv
(
service_id int,
team_id int
);
INSERT INTO tbl_team_list_serv
VALUES
(7, 2),
(9, 3),
(10, 4);
CREATE TABLE tbl_service
(
id int,
parent int
);
INSERT INTO tbl_service
VALUES
(5, null),
(6, 5),
(7, 6),
(8, null),
(9, 8),
(10, null);
为了简单起见,我声明:
1
作为 employee_id
2, 3, 4
作为 team_id
5 -> 6 -> 7
作为服务(5为主服务)
8 -> 9
(8为主要服务)
10
(10为主要服务)
检索员工所属的服务我查询
SELECT ls.service_id FROM tbl_team_list_serv ls
JOIN tbl_employee_team t ON ls.team_id=t.teams_id WHERE t.employee_id = 1
从我使用的服务中获取主要服务
WITH RECURSIVE r AS
(
SELECT id, parent, 1 AS level
FROM tbl_service
WHERE id = 7 /*(here's I need to assign to every id from the JOIN)*/
UNION
SELECT tbl_service.id, tbl_service.parent, r.level + 1 AS level
FROM tbl_service
JOIN r
ON r.parent = tbl_service.id
)
SELECT id FROM r WHERE r.level = (SELECT max(level) FROM r)
我的问题是如何合并这两个查询?
根据上面的数据,我想最终得到一个 id 列表,在这种情况下是这样的:
5, 8, 10
此外,我希望我的递归查询到 return 最后一行(我不认为带有 level 的解决方案很优雅)
可以找到 SQLFiddle here
提前致谢
我觉得你已经完成了这个问题的大部分工作。这只是以下调整的问题:
- 将第一个查询的逻辑放在 CTE 的锚点部分。
- 将原始服务 ID 添加为一列以记住层次结构。
- 调整最终逻辑,使每个原始服务获得一行。
作为查询:
WITH RECURSIVE r AS (
SELECT ls.service_id as id, s.parent, 1 as level, ls.service_id as orig_service_id
FROM tbl_team_list_serv ls JOIN
tbl_employee_team t
ON ls.team_id = t.teams_id JOIN
tbl_service s
ON ls.service_id = s.id
WHERE t.employee_id = 1
UNION ALL
SELECT s.id, s.parent, r.level + 1 AS level, r.orig_service_id
FROM tbl_service s JOIN
r
ON r.parent = s.id
)
SELECT r.id
FROM (SELECT r.*,
MAX(level) OVER (PARTITION BY orig_service_id) as max_level
FROM r
) r
WHERE r.level = max_level;
Here 是一个 db<>fiddle.
我有以下架构:
CREATE TABLE tbl_employee_team
(
employee_id int,
teams_id int
);
INSERT INTO tbl_employee_team
VALUES
(1, 2),
(1, 3),
(1, 4);
CREATE TABLE tbl_team_list_serv
(
service_id int,
team_id int
);
INSERT INTO tbl_team_list_serv
VALUES
(7, 2),
(9, 3),
(10, 4);
CREATE TABLE tbl_service
(
id int,
parent int
);
INSERT INTO tbl_service
VALUES
(5, null),
(6, 5),
(7, 6),
(8, null),
(9, 8),
(10, null);
为了简单起见,我声明:
1
作为 employee_id
2, 3, 4
作为 team_id
5 -> 6 -> 7
作为服务(5为主服务)
8 -> 9
(8为主要服务)
10
(10为主要服务)
检索员工所属的服务我查询
SELECT ls.service_id FROM tbl_team_list_serv ls
JOIN tbl_employee_team t ON ls.team_id=t.teams_id WHERE t.employee_id = 1
从我使用的服务中获取主要服务
WITH RECURSIVE r AS
(
SELECT id, parent, 1 AS level
FROM tbl_service
WHERE id = 7 /*(here's I need to assign to every id from the JOIN)*/
UNION
SELECT tbl_service.id, tbl_service.parent, r.level + 1 AS level
FROM tbl_service
JOIN r
ON r.parent = tbl_service.id
)
SELECT id FROM r WHERE r.level = (SELECT max(level) FROM r)
我的问题是如何合并这两个查询?
根据上面的数据,我想最终得到一个 id 列表,在这种情况下是这样的:
5, 8, 10
此外,我希望我的递归查询到 return 最后一行(我不认为带有 level 的解决方案很优雅)
可以找到 SQLFiddle here
提前致谢
我觉得你已经完成了这个问题的大部分工作。这只是以下调整的问题:
- 将第一个查询的逻辑放在 CTE 的锚点部分。
- 将原始服务 ID 添加为一列以记住层次结构。
- 调整最终逻辑,使每个原始服务获得一行。
作为查询:
WITH RECURSIVE r AS (
SELECT ls.service_id as id, s.parent, 1 as level, ls.service_id as orig_service_id
FROM tbl_team_list_serv ls JOIN
tbl_employee_team t
ON ls.team_id = t.teams_id JOIN
tbl_service s
ON ls.service_id = s.id
WHERE t.employee_id = 1
UNION ALL
SELECT s.id, s.parent, r.level + 1 AS level, r.orig_service_id
FROM tbl_service s JOIN
r
ON r.parent = s.id
)
SELECT r.id
FROM (SELECT r.*,
MAX(level) OVER (PARTITION BY orig_service_id) as max_level
FROM r
) r
WHERE r.level = max_level;
Here 是一个 db<>fiddle.