SQL开发知识：详细聊一聊mysql的树形结构存储以及查询

CREATE TABLE `menu` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(50) NOT NULL,
`parent_id` int(11) NOT NULL DEFAULT ‘0’,
PRIMARY KEY (`id`)
) ENGINE=InnoDB;

INSERT INTO `menu` (`id`, `name`, `parent_id`) VALUES
(1, ‘level1a’, 0),
(2, ‘level1b’, 0),
(3, ‘level2a-1a’,1),
(4, ‘level2b-1a’,1),
(5, ‘level2a-1b’, 2),
(6, ‘level2b-1b’, 2),
(7, ‘level3-2a1a’, 3),
(8, ‘level3-2b1a’, 4),
(9, ‘level3-2a1b’, 5),
(10, ‘level3-2b1b’, 6);

— 查询跟节点下的所有节点
SELECT t1.name AS lev1, t2.name as lev2, t3.name as lev3
FROM menu AS t1
LEFT JOIN menu AS t2 ON t2.parent_id = t1.id
LEFT JOIN menu AS t3 ON t3.parent_id = t2.id
WHERE t1.name = ‘level1a’;

+———+————+————-+
| lev1 | lev2 | lev3 |
+———+————+————-+
| level1a | level2a-1a | level3-2a1a |
| level1a | level2b-1a | level3-2b1a |
+———+————+————-+

— 查询叶子节点
SELECT t1.name FROM
menu AS t1 LEFT JOIN menu as t2
ON t1.id = t2.parent_id
WHERE t2.id IS NULL;

+————-+
| name |
+————-+
| level3-2a1a |
| level3-2b1a |
| level3-2a1b |
| level3-2b1b |
+————-+

CREATE TABLE `menu_path` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(50) NOT NULL,
`parent_id` int(11) NOT NULL DEFAULT ‘0’,
`path` varchar(255) NOT NULL DEFAULT ”,
PRIMARY KEY (`id`)
) ENGINE=InnoDB;

INSERT INTO `menu_path` (`id`, `name`, `parent_id`, `path`) VALUES
(1, ‘level1a’, 0, ‘1/’),
(2, ‘level1b’, 0, ‘2/’),
(3, ‘level2a-1a’,1, ‘1/3’),
(4, ‘level2b-1a’,1, ‘1/4’),
(5, ‘level2a-1b’, 2, ‘2/5’),
(6, ‘level2b-1b’, 2, ‘2/6’),
(7, ‘level3-2a1a’, 3, ‘1/3/7’),
(8, ‘level3-2b1a’, 4, ‘1/4/8’),
(9, ‘level3-2a1b’, 5, ‘2/5/9’),
(10, ‘level3-2b1b’, 6, ‘2/6/10’);

— 查询某个节点的所有子节点
select * from menu_path where path like ‘1/%’
+—-+————-+———–+——-+
| id | name | parent_id | path |
+—-+————-+———–+——-+
| 1 | level1a | 0 | 1/ |
| 3 | level2a-1a | 1 | 1/3 |
| 4 | level2b-1a | 1 | 1/4 |
| 7 | level3-2a1a | 3 | 1/3/7 |
| 8 | level3-2b1a | 4 | 1/4/8 |
+—-+————-+———–+——-+

CREATE TABLE `menu_preorder` (
`id` int(11) NOT NULL,
`name` varchar(50) NOT NULL,
`lft` int(11) NOT NULL DEFAULT ‘0’,
`rgt` int(11) NOT NULL DEFAULT ‘0’,
PRIMARY KEY (`id`)
) ENGINE=InnoDB;

1(level1a)14
2(level2a)7 8(level2b)13
3(level3a-2a)4 5(level3b-2a)6 9(level3c-2b)10 11(level3d-2b)12

INSERT INTO `menu_preorder` (`id`, `name`, `lft`, `rgt`) VALUES
(1, ‘level1a’, 1, 14),
(2, ‘level2a’,2, 7),
(3, ‘level2b’,8, 13),
(4, ‘level3a-2a’, 3, 4),
(5, ‘level3b-2a’, 5, 6),
(6, ‘level3c-2b’, 9, 10),
(7, ‘level3d-2b’, 11, 12);

select * from menu_preorder
+—-+————+—–+—–+
| id | name | lft | rgt |
+—-+————+—–+—–+
| 1 | level1a | 1 | 14 |
| 2 | level2a | 2 | 7 |
| 3 | level2b | 8 | 13 |
| 4 | level3a-2a | 3 | 4 |
| 5 | level3b-2a | 5 | 6 |
| 6 | level3c-2b | 9 | 10 |
| 7 | level3d-2b | 11 | 12 |
+—-+————+—–+—–+

— 查询某个节点及其子节点，比如level2b
select * from menu_preorder where lft between 8 and 13
+—-+————+—–+—–+
| id | name | lft | rgt |
+—-+————+—–+—–+
| 3 | level2b | 8 | 13 |
| 6 | level3c-2b | 9 | 10 |
| 7 | level3d-2b | 11 | 12 |
+—-+————+—–+—–+

— 查询所有叶子节点
SELECT name
FROM menu_preorder
WHERE rgt = lft + 1;

+————+
| name |
+————+
| level3a-2a |
| level3b-2a |
| level3c-2b |
| level3d-2b |
+————+

— 查询某个节点及其父节点
SELECT parent.*
FROM menu_preorder AS node,
menu_preorder AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
AND node.name = ‘level2b’
ORDER BY parent.lft;

+—-+———+—–+—–+
| id | name | lft | rgt |
+—-+———+—–+—–+
| 1 | level1a | 1 | 14 |
| 3 | level2b | 8 | 13 |
+—-+———+—–+—–+

— 树形结构展示
SELECT CONCAT( REPEAT(‘ ‘, COUNT(parent.name) – 1), node.name) AS name
FROM menu_preorder AS node,
menu_preorder AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY node.name
ORDER BY node.lft;

+————–+
| name |
+————–+
| level1a |
| level2a |
| level3a-2a |
| level3b-2a |
| level2b |
| level3c-2b |
| level3d-2b |
+————–+