在父子层次结构中查找最多 7 个级别的级别段

Question

父子表

id	introducer_id	name
1	NULL	Riya
2	1	Ramesh
3	1	Anand
4	2	Preety
5	3	Rakesh

id = 1时查询成员列表

select  id AS memberid, name 
from (select * 
      from table_member 
      order by introducer_id, id) table_member_sorted,
     (select @pv := '1') initialisation 
where find_in_set(introducer_id, @pv) 
  and length(@pv := concat(@pv, ',', id))

以上查询显示输出

member_id	name
2	Ramesh
3	Anand
4	Preety
5	Rakesh

现在我也想查询所有成员的等级 （即）我希望输出为：

member_id	name	level
2	Ramesh	1
3	Anand	1
4	Preety	2
5	Rakesh	2

如何更新我之前的查询以获得上述输出？

注意：查询最多显示7级

谢谢。

Answer 1

CREATE PROCEDURE get_tree (start_from INT)
BEGIN
    DROP TABLE IF EXISTS tmp;
    CREATE TABLE tmp (member_id INT PRIMARY KEY, 
                      name VARCHAR(255),
                      level INT);
    INSERT INTO tmp SELECT id, name, 0
                    FROM test
                    WHERE id = start_from;
    REPEAT
        INSERT IGNORE INTO tmp SELECT test.id, test.name, tmp.level + 1
                               FROM test
                               JOIN tmp ON tmp.member_id = test.introducer_id;
    UNTIL !ROW_COUNT() END REPEAT;
    SELECT * FROM tmp;
END

fiddle

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我已经分享了我的 sql 版本的截图链接。 snipboard.io/qbdB0A.jpg – AfreenB

您的 SQL 服务器版本是 MariaDB 10.4.14。 ——秋名

为您的服务器版本使用

WITH RECURSIVE
cte AS ( SELECT id member_id, name, 0 level
         FROM test
         WHERE id = @start_from 
       UNION ALL
         SELECT test.id, test.name, cte.level + 1
         FROM test
         JOIN cte ON cte.member_id = test.introducer_id 
      -- WHERE cte.level < 7 )
SELECT * 
FROM cte;

fiddle