我在Postgres 9.6中有一个接受int[]
参数的函数。我想让该函数也接受单个int(并在必要时将其转换为单个元素数组)。
CREATE OR REPLACE FUNCTION get_subordinates(inp_persona_ids integer[])
-- Get all subordnates of the people passed in as array
-- TODO allow a single persona ID (int) to be passed in as inp_persona_ids
RETURNS TABLE (persona_id int) AS
$$
BEGIN
RETURN QUERY(
WITH RECURSIVE children AS (
-- passed in persona_id
SELECT
id AS persona_id,
manager_id
FROM
personas
WHERE
id = ANY(inp_persona_ids)
UNION
-- and all subordinates
SELECT
p.id AS persona_id,
p.manager_id
FROM
personas p
JOIN children c ON p.manager_id = c.persona_id
)
SELECT
children.persona_id
FROM
children
LEFT JOIN
personas on children.persona_id = personas.id
WHERE personas.disabled IS NOT TRUE
);
END;
$$ LANGUAGE plpgsql
我将如何更改函数定义,并添加一些条件逻辑来测试int
,并在必要时更改为ARRAY[int]
?
不可能在单个函数中处理,但是您可以使用integer
参数重载该函数并将其作为数组传递给现有函数:
CREATE OR REPLACE FUNCTION get_subordinates(inp_persona_id integer)
RETURNS TABLE (persona_id int) AS
$$
BEGIN
RETURN QUERY SELECT * FROM get_subordinates(ARRAY[inp_persona_id]);
END;
$$ LANGUAGE plpgsql;
也许您可能还想根据NULL
检查参数,这取决于您。
使用VARIADIC
修饰符使用单个功能可以实现[[<< [is:
CREATE OR REPLACE FUNCTION get_subordinates(VARIADIC inp_persona_ids int[])
RETURNS TABLE (persona_id int) AS
$func$
WITH RECURSIVE children AS ( -- passed in persona_id
SELECT id AS persona_id, manager_id, disabled
FROM personas
WHERE id = ANY(inp_persona_ids)
UNION ALL -- and all subordinates
SELECT p.id AS persona_id
, p.manager_id
FROM children c
JOIN personas p ON p.manager_id = c.persona_id
)
SELECT c.persona_id
FROM children c
WHERE c.disabled IS NOT TRUE
$func$ LANGUAGE sql;
但是在提供数组而不是列表时,需要在调用中添加关键字
VARIADIC
:SELECT * FROM get_subordinates(VARIADIC '{1,2,3}'::int[]);
SELECT * FROM get_subordinates(1,2,3);
SELECT * FROM get_subordinates(1);
如果这不是一个选择,您将按照另一个答案中的建议返回功能重载。
见:
UNION
毫无意义。仅当您的树绕圈旋转时才会发生重复,这将导致无限循环,并且rCTE会出错。使用便宜的UNION ALL
。LEFT JOIN
毫无意义。无论如何,添加的WHERE
强制其表现得像普通的[INNER] JOIN
。disabled
。