选择[COLUMN_NAME] AS，自引用表

Question

下面是我在自引用表get_reportees上执行的函数emp_tabref1

CREATE OR REPLACE FUNCTION get_reportees4(IN id integer)
RETURNS TABLE(e_id integer, e_name character varying, e_manager integer, e_man_name character varying) AS
$$
BEGIN
RETURN QUERY
WITH RECURSIVE manger_hierarchy(e_id, e_name, m_id, m_name) AS 
(
SELECT e.emp_id, e.emp_name, e.mgr_id, e.emp_name AS man_name
FROM emp_tabref1 e WHERE e.emp_id = id
UNION
SELECT rp.emp_id, rp.emp_name, rp.mgr_id, rp.emp_name AS man_name
FROM manger_hierarchy mh INNER JOIN emp_tabref1 rp ON mh.e_id = rp.mgr_id
)
SELECT * from manger_hierarchy;
END;
$$ LANGUAGE plpgsql VOLATILE

Emp_tabref1的表结构：

CREATE TABLE **emp_tabref1**
(
emp_id integer NOT NULL,
emp_name character varying(50) NOT NULL,
mgr_id integer,
CONSTRAINT emp_tabref_pkey PRIMARY KEY (emp_id),
CONSTRAINT emp_tabref_mgr_id_fkey FOREIGN KEY (mgr_id)
  REFERENCES emp_tabref (emp_id) MATCH SIMPLE
  ON UPDATE NO ACTION ON DELETE NO ACTION
)

我希望返回的是我们要传递的id的层次结构(包括上面和下面)，它将具有emp_name、emp_id、mgr_id和mgr_name。

但我的功能是这样回来的：

select * from get_reportees4(9)

   e_id e_name  e_manager e_man_name
1    9  "Emp9"  10        "Emp9"
2    5  "Emp5"  9         "Emp5"
3    6  "Emp6"  9         "Emp6"

我的预期产出在哪里

       e_id e_name  e_manager e_man_name
1    9  "Emp9"        10        "Emp10"
2    5  "Emp5"         9        "Emp9"
3    6  "Emp6"         9        "Emp9"

函数应该返回经理名，而不是员工名。请帮帮我！

Answer 1

找到解决办法了！通过在临时manger_hierarchy表和emp_tabref1表之间使用mgr_id和emp_id创建一个新的连接

CREATE OR REPLACE FUNCTION get_reportees4(IN id integer)
RETURNS TABLE(e_id integer, e_name character varying, e_manager integer, e_man_name character varying) AS
$$
BEGIN
RETURN QUERY 
WITH RECURSIVE manger_hierarchy(e_id, e_name, m_id, m_name) AS 
(
SELECT e.emp_id, e.emp_name, e.mgr_id, e.emp_name AS man_name
FROM emp_tabref1 e WHERE e.emp_id = id
UNION
SELECT rp.emp_id, rp.emp_name, rp.mgr_id, rp.emp_name AS man_name
FROM manger_hierarchy mh INNER JOIN emp_tabref1 rp ON mh.e_id = rp.mgr_id 
)
SELECT manger_hierarchy.e_id, manger_hierarchy.e_name, manger_hierarchy.m_id, emp_tabref1.emp_name 
FROM manger_hierarchy LEFT JOIN emp_tabref1 ON manger_hierarchy.m_id = emp_tabref1.emp_id;
END;
$$ LANGUAGE plpgsql VOLATILE

选择manger_hierarchy.e_id，manger_hierarchy.e_name，manger_hierarchy.m_id，emp_tabref1.emp_name从manger_hierarchy左加入emp_tabref1 ON manger_hierarchy.m_id = emp_tabref1.emp_id;