Sqlite 限制具有相同 id 的重复行

Question

我正在使用 android SQLite 数据库，结果如下：

 id | root_id | parent_id   |name
----------------------------------
613 | null    | null        | m1
612 | null    | null        | m4
570 | null    | null        | m1
635 | 570     | 570         | m6
653 | 570     | 635         | m1
652 | 570     | 635         | m3
632 | 570     | 570         | m9
392 | null    | null        | m2
753 | 392     | 392         | m5
751 | 392     | 392         | m4
391 | null    | null        | m7

我通过下面的查询得到这个结果：

WITH RECURSIVE all_employees(id, root_id, parent_id, name) AS (
    SELECT id, id AS root_id, parent_id, name FROM employees WHERE parent_id IS NULL
    UNION ALL
    SELECT c.id, (CASE WHEN c.parent_id IS NOT NULL THEN p.root_id END), c.parent_id, c.name FROM employees c JOIN all_employees p ON p.id = c.parent_id ORDER BY id DESC
)

SELECT id, (CASE WHEN id != root_id THEN root_id ELSE NULL END) root_id, parent_id, name FROM all_employees

我想为具有相同 root_id 的重复行设置一个限制，例如： 加载 2 行 root_id 570 和 1 行 root_id 392：

 id | root_id | parent_id   |name
----------------------------------
613 | null    | null        | m1
612 | null    | null        | m4
570 | null    | null        | m1
635 | 570     | 570         | m6
653 | 570     | 635         | m1
392 | null    | null        | m2
753 | 392     | 392         | m5
391 | null    | null        | m7

Answer 1

创建另一个CTE：

counters(root_id, n) AS (VALUES (570, 2), (392, 1)) 

您返回所有要限制其行的root_ids 以及每行的行数，然后使用递归CTE 的LEFT 连接。
最后，在WHERE 子句中使用相关子查询设置条件：

WITH 
  RECURSIVE all_employees(id, root_id, parent_id, name) AS (
    SELECT id, id AS root_id, parent_id, name 
    FROM employees 
    WHERE parent_id IS NULL
    UNION ALL
    SELECT c.id, (CASE WHEN c.parent_id IS NOT NULL THEN p.root_id END), c.parent_id, c.name 
    FROM employees c JOIN all_employees p 
    ON p.id = c.parent_id 
  ),
  counters(root_id, n) AS (VALUES (570, 2), (392, 1))
SELECT a.id, (CASE WHEN a.id <> a.root_id THEN a.root_id END) root_id, a.parent_id, a.name
FROM all_employees a LEFT JOIN counters c
ON c.root_id = a.root_id
WHERE c.root_id IS NULL 
   OR (SELECT COUNT(*) FROM all_employees b WHERE b.root_id IS a.root_id AND b.id <= a.id) <= c.n

请注意，CTE 中的 ORDER BY 子句是没有用的，因为当您从 CTE 中选择时，不能保证行将按该顺序返回。
您可以在最终查询中设置所需行的顺序。

Answer 2

有限选择的联合

WITH RECURSIVE all_employees(id, root_id, parent_id, name) AS (
    SELECT id, id AS root_id, parent_id, name 
    FROM employees 
    WHERE parent_id IS NULL
    UNION ALL
    SELECT c.id, (CASE WHEN c.parent_id IS NOT NULL THEN p.root_id END), c.parent_id, c.name 
    FROM employees c 
    JOIN all_employees p ON p.id = c.parent_id ORDER BY id DESC
)
SELECT id, root_id, parent_id, name
FROM all_employees
where root_id not in (570, 392)
union all
select *
from (
   select *
   from all_employees
   where root_id =570 
   limit 3
)
union all
select *
from (
   select *
   from all_employees
   where root_id =392
   limit 2
)

Answer 3

向 CTE 添加一个计数器，然后使用WHERE 逻辑进行过滤：

WITH RECURSIVE all_employees(id, root_id, parent_id, name, lev) AS (
    SELECT id, id AS root_id, parent_id, name, 0 as lev
    FROM employees
    WHERE parent_id IS NULL
    UNION ALL
    SELECT c.id,
           (CASE WHEN c.parent_id IS NOT NULL THEN p.root_id END), c.parent_id, c.name, p.lev + 1
    FROM employees c JOIN
         all_employees p
         ON p.id = c.parent_id
    ORDER BY id DESC   -- don't think this does anything
   )
SELECT id, (CASE WHEN id <> root_id THEN root_id END) as root_id, parent_id, name
FROM all_employees
WHERE root_id = 570 AND cnt <= 1 OR
      root_id = 392 AND cnt <= 2 OR
      root_id NOT IN (570, 392);

这里的重点是将计数放入递归 CTE 中，然后使用该信息进行过滤。您可以随意表达WHERE 子句：

WHERE (CASE WHEN root_id = 570 THEN cnt <= 1
            WHEN root_id = 392 THEN cnt <= 2
            ELSE 1
       END)