SQL 查询到 return 匹配的行及其树状结构中的祖先

SQL query to return matched rows with their ancestors in tree-like structure

我有一个 Oracle 12c 数据库,其中 table 具有组织单位(部门等)的树状结构:

CREATE TABLE "OUS" (
  "ID" NUMBER(38,0) NOT NULL ENABLE, 
  "NAME" VARCHAR2(255 CHAR) NOT NULL ENABLE, 
  "PARENT_ID" NUMBER(38,0),
  PRIMARY KEY("ID"),
  CONSTRAINT "OUS_HIERARCHY_FK" FOREIGN KEY ("PARENT_ID") REFERENCES "OUS" ("ID") ON DELETE CASCADE
);

所以,结构像

| id | name          | parent_id |
| -: | ------------- | --------: |
|  1 | Root          |    (NULL) |
|  2 | Territorial 1 |         1 |
|  3 | Regional 1-1  |         2 |
|  4 | Alpha dept    |         3 |
|  5 | Beta dept     |         3 |
|  6 | Regional 1-2  |         2 |
|  7 | Gamma dept    |         6 |
|  8 | Delta dept    |         7 |
|  9 | Territorial 2 |         1 |
| 10 | Regional 2-1  |         9 |
| 11 | Epsilon dept  |        10 |
| 12 | Zeta dept     |        10 |

您可以使用 SQL 创建它,例如:

INSERT INTO ous (id, name, parent_id) VALUES ( 13, 'Root',          NULL);
INSERT INTO ous (id, name, parent_id) VALUES (  2, 'Territorial 1',   13);
INSERT INTO ous (id, name, parent_id) VALUES (  1, 'Regional 1-1',     2);
INSERT INTO ous (id, name, parent_id) VALUES (  5, 'Alpha dept',       1);
INSERT INTO ous (id, name, parent_id) VALUES (  4, 'Beta dept',        1);
INSERT INTO ous (id, name, parent_id) VALUES (  6, 'Regional 1-2',     2);
INSERT INTO ous (id, name, parent_id) VALUES (  7, 'Gamma dept',       6);
INSERT INTO ous (id, name, parent_id) VALUES (  8, 'Delta dept',       6);
INSERT INTO ous (id, name, parent_id) VALUES (  9, 'Territorial 2',   13);
INSERT INTO ous (id, name, parent_id) VALUES (  3, 'Regional 2-1',     9);
INSERT INTO ous (id, name, parent_id) VALUES ( 15, 'Epsilon dept',     3);
INSERT INTO ous (id, name, parent_id) VALUES ( 12, 'Zeta dept',        3);

我需要找到一些 OU,匹配给定的条件(如 name = 'Alpha' OR name = 'Epsilon)并获得这些 OU 及其祖先的子树。

例如:

| id | name          | parent_id |
| -: | ------------- | --------: |
|  1 | Root          |    (NULL) | ← Ancestor of Alpha and Epsilon
|  2 | Territorial 1 |         1 | ← Ancestor of Alpha
|  3 | Regional 1-1  |         2 | ← Ancestor of Alpha
|  4 | Alpha dept    |         3 | ← Matches the WHERE clause!
|  9 | Territorial 2 |         1 | ← Ancestor of Epsilon
| 10 | Regional 2-1  |         9 | ← Ancestor of Epsilon
| 11 | Epsilon dept  |        10 | ← Matches the WHERE clause!

我查看了各种 Hierarchical and recursive queries in SQL: Oracle Hierarchical queries and CTEs,但无法找出可以 return 我这样的结果的查询。

我正在使用 Oracle Database 12c。

我试过这样的查询:

SELECT ous.* FROM ous
WHERE name = 'Alpha' OR name = 'Epsilon'
START WITH 
  parent_id IS NULL
CONNECT BY
  PRIOR id = parent_id 
ORDER SIBLINGS BY name;

但是它 returns 0 行,因为 WHERE 应用于所有行(所以祖先被过滤)

我也试过了:

WITH RECURSIVE all_nodes (id, parent_id, name) AS (
  SELECT ous.id, ous.parent_id, name FROM ous WHERE (name = 'Alpha' OR name = 'Epsilon')
  UNION
  SELECT ous.id, ous.parent_id, name FROM ous INNER JOIN all_nodes ON ous.parent_id = all_nodes.id
)
SELECT * FROM all_nodes INNER JOIN ous ON all_nodes.id = ous.id ORDER BY name;

但是 return 错误 SQL Error [905] [42000]: ORA-00905: keyword is missing

您可以使用递归 CTE 执行此操作:

with t(name, id, parent_id) as (
      select name, id, parent_id
      from ous
      where name in ('alpha', 'epsilon')
      union all
      select ous.name, ous.id, ous.parent_id
      from t join
           ous
           on ous.id = t.parent_id
    )
select distinct t.id, t.name, t.parent
from t
order by t.id;

可能不需要select distinct

递归 CTE 具有标准 SQL 的优势,因此许多不同的数据库都支持该逻辑。​​

当然,您可以为此使用分层查询。问题是,如果你从多个叶子开始,在某个时候你会开始得到重复的行。您可以使用 "distinct" 删除重复项,但这会影响性能,尤其是在非常大的 table 上或者如果您开始时有太多叶子。归根结底,递归查询通常更难编写,但比分层查询更高效。

为了完整起见,这里是分层解决方案。使用 SCOTT 架构中的 EMP table 进行说明。首先显示直接分层查询(以及输出中的重复项),然后显示带有 "distinct."

的版本
select empno, mgr
from   scott.emp
start with empno in (7902, 7788)
connect by prior mgr = empno
;

     EMPNO        MGR
---------- ----------
      7788       7566
      7566       7839
      7839
      7902       7566
      7566       7839
      7839


select distinct empno, mgr
from   scott.emp
start with empno in (7902, 7788)
connect by prior mgr = empno
;

     EMPNO        MGR
---------- ----------
      7839
      7566       7839
      7902       7566
      7788       7566