MySQL多层级结构-树搜索介绍

CREATE TABLE acos (
 id INTEGER(10) UNSIGNED NOT NULL AUTO_INCREMENT,
 parent_id INTEGER(10) DEFAULT NULL,
 model VARCHAR(255) DEFAULT '',
 foreign_key INTEGER(10) UNSIGNED DEFAULT NULL,
 alias VARCHAR(255) DEFAULT '',
 lft INTEGER(10) DEFAULT NULL,
 rght INTEGER(10) DEFAULT NULL,
 PRIMARY KEY (id)
);

DROP TABLE IF EXISTS comment;
CREATE TABLE `comment` (
 `comment_id` int(11) DEFAULT NULL,
 `left_num` int(11) DEFAULT NULL,
 `right_num` int(11) DEFAULT NULL
);

INSERT INTO `comment` VALUES 
 (1,1,14),
 (2,2,5),
 (3,3,4),
 (4,6,13),
 (5,7,8),
 (6,9,12),
 (7,10,11);
 
CREATE INDEX idx$comment$left_num$right_num ON `comment` (`left_num`, `right_num`);

-- 获得 '节点4' 孩子
SELECT c.*
FROM comment AS p, comment AS c
WHERE c.left_num BETWEEN p.left_num AND p.right_num
 AND p.comment_id = 4;
+------------+----------+-----------+
| comment_id | left_num | right_num |
+------------+----------+-----------+
|     4 |    6 |    13 |
|     5 |    7 |     8 |
|     6 |    9 |    12 |
|     7 |    10 |    11 |
+------------+----------+-----------+

-- 获得 '节点6' 父亲
SELECT p.* 
FROM comment AS p, comment AS c
WHERE c.left_num BETWEEN p.left_num AND p.right_num
 AND c.comment_id = 6;
+------------+----------+-----------+
| comment_id | left_num | right_num |
+------------+----------+-----------+
|     1 |    1 |    14 |
|     4 |    6 |    13 |
|     6 |    9 |    12 |
+------------+----------+-----------+

SET SESSION sql_mode = 'STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUBSTITUTION';
SELECT c.*,
 COUNT(c.comment_id) AS depth
FROM comment AS p, comment AS c
WHERE c.left_num BETWEEN p.left_num AND p.right_num
 AND c.comment_id = 4
GROUP BY c.comment_id;
+------------+----------+-----------+-------+
| comment_id | left_num | right_num | depth |
+------------+----------+-----------+-------+
|     4 |    6 |    13 |   2 |
+------------+----------+-----------+-------+

SELECT sub_child.*,
 (COUNT(sub_parent.comment_id) - 1) AS depth
FROM (
 SELECT child.*
 FROM comment AS parent, comment AS child
 WHERE child.left_num BETWEEN parent.left_num AND parent.right_num
  AND parent.comment_id = 4
) AS sub_child, (
 SELECT child.*
 FROM comment AS parent, comment AS child
 WHERE child.left_num BETWEEN parent.left_num AND parent.right_num
  AND parent.comment_id = 4
) AS sub_parent
WHERE sub_child.left_num BETWEEN sub_parent.left_num AND sub_parent.right_num
GROUP BY sub_child.comment_id
ORDER BY sub_child.left_num;
+------------+----------+-----------+-------+
| comment_id | left_num | right_num | depth |
+------------+----------+-----------+-------+
|     4 |    6 |    13 |   0 |
|     5 |    7 |     8 |   1 |
|     6 |    9 |    12 |   1 |
|     7 |    10 |    11 |   2 |
+------------+----------+-----------+-------+

-- 获得 '节点4' 和 '节点4'的第一个孩子的(节点5)的左右值
SELECT c.*
FROM comment AS p, comment AS c
WHERE c.left_num BETWEEN p.left_num AND p.right_num
 AND p.comment_id = 4;
+------------+----------+-----------+
| comment_id | left_num | right_num |
+------------+----------+-----------+
|     4 |    6 |    13 |
|     5 |    7 |     8 |
... omit ...
-- 通过上面获得的信息更新 '节点4' 的父子几点的左右值
UPDATE comment SET left_num = left_num + 2 WHERE left_num > 6;
UPDATE comment SET right_num = right_num + 2 WHERE right_num > 6;

INSERT INTO comment 
SELECT 44, left_num + 1, left_num + 2
FROM comment WHERE comment_id = 4;

-- 获得 '节点4' 孩子
SELECT c.*
FROM comment AS p, comment AS c
WHERE c.left_num BETWEEN p.left_num AND p.right_num
 AND p.comment_id = 4;
+------------+----------+-----------+
| comment_id | left_num | right_num |
+------------+----------+-----------+
|     4 |    6 |    15 |
|     5 |    9 |    10 |
|     6 |    11 |    14 |
|     7 |    12 |    13 |
|     44 |    7 |     8 |
+------------+----------+-----------+
-- 获得 '节点44' 父亲
SELECT p.* 
FROM comment AS p, comment AS c
WHERE c.left_num BETWEEN p.left_num AND p.right_num
 AND c.comment_id = 44;
+------------+----------+-----------+
| comment_id | left_num | right_num |
+------------+----------+-----------+
|     1 |    1 |    16 |
|     4 |    6 |    15 |
|     44 |    7 |     8 |
+------------+----------+-----------+