MySQL 8.0 版 架构 SQL
CREATE TABLE IF NOT EXISTS `department` (
`id` INT NOT NULL,
`name` VARCHAR(45) NOT NULL,
`father` INT NULL,
PRIMARY KEY (`id`),
INDEX `fk_department_department_idx` (`father` ASC) VISIBLE,
CONSTRAINT `fk_department_department`
FOREIGN KEY (`father`)
REFERENCES `department` (`id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB;
insert into department (id,name,father)
values
(1, 'dp1',null),
(2, 'dp2',null),
(3, 'dp3',1),
(4, 'dp4',1),
(5, 'dp5',2),
(6, 'dp6',4),
(7, 'dp7',6),
(8, 'dp8',6),
(9, 'dp9',6);
SET GLOBAL sql_mode=(SELECT REPLACE(@@sql_mode,'ONLY_FULL_GROUP_BY',''));
SET SESSION sql_mode = '';
我的查询:
WITH RECURSIVE cte_department AS (
SELECT
d1.id,
d1.name,
d1.father
FROM
department d1
WHERE
d1.id=6
UNION ALL
SELECT
d2.id,
d2.name,
d2.father
FROM
department d2
INNER JOIN cte_department cte ON cte.id = d2.father
)
SELECT * FROM cte_department;
结果:
id name father
6 dp6 4
7 dp7 6
8 dp8 6
9 dp9 6
我需要什么:
id name father
1 dp1 null
4 dp4 1
6 dp6 4
7 dp7 6
8 dp8 6
9 dp9 6
问题是: 我可以获取所有 child ,但我需要将给定 ID 中的所有 parent 添加到此查询中,在本例中为 ID 6。 我坚持这一点。如果有人可以帮助我,请跟随 fiddle 。
最佳答案
递归部分可以有多个查询 block 。
WITH RECURSIVE cte_department AS (
SELECT
d1.id,
d1.name,
d1.father,
'Begin' state
FROM
department d1
WHERE
d1.id=6
UNION ALL
SELECT
d2.id,
d2.name,
d2.father,
'Up'
FROM
department d2
INNER JOIN
cte_department cte
ON
cte.father = d2.id
WHERE
cte.state in ('Begin', 'Up')
UNION ALL
SELECT
d2.id,
d2.name,
d2.father,
'Down'
FROM
department d2
INNER JOIN
cte_department cte
ON
cte.id = d2.father
WHERE
cte.state in ('Begin', 'Down')
)
SELECT
id, name, father
FROM
cte_department
ORDER BY
father, id, name;
试用 db<>fiddle .
关于MySql Recursive - 从给定的id获取所有 child 和 parent ,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/65377583/