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首页 > 解决方案 > 在我的 sql < 8 中获取树文件夹结构的父母和孩子，并且没有 CTE

mysql - 在我的 sql < 8 中获取树文件夹结构的父母和孩子，并且没有 CTE

问题描述

id我有一个文件夹表，它以,parent_id关系连接到自身：

CREATE TABLE folders (
  id int(10) unsigned NOT NULL AUTO_INCREMENT,
  title nvarchar(255) NOT NULL,
  parent_id int(10) unsigned DEFAULT NULL,
  PRIMARY KEY (id)
);

INSERT INTO folders(id, title, parent_id) VALUES(1, 'root', null);
INSERT INTO folders(id, title, parent_id) values(2, 'one', 1);
INSERT INTO folders(id, title, parent_id) values(3, 'target', 2);
INSERT INTO folders(id, title, parent_id) values(4, 'child one', 3);
INSERT INTO folders(id, title, parent_id) values(5, 'child two', 3);
INSERT INTO folders(id, title, parent_id) values(6, 'root 2', null);
INSERT INTO folders(id, title, parent_id) values(7, 'other child one', 6);
INSERT INTO folders(id, title, parent_id) values(8, 'other child two', 6);

我想要一个查询，它返回该记录的所有父母，直接回到路线和任何孩子。

因此，如果我要求带有 的文件夹id=3，我会得到记录：1, 2, 3, 4, 5。我很困惑如何得到父母。

MYSQL 的版本是 5.7，并且没有立即升级的计划，所以遗憾的是 CTE 不是一个选项。

我已经创建了这个sql fiddle

标签： mysqlsqlhierarchical-datarecursive-query

解决方案

在您的表设计中，ID对应PARENT_ID于“邻接表模型”，用于存储一棵树。

还有另一种设计，称为“嵌套集模型”，它可以更轻松地在此处执行您想要的操作。

请参阅 Mike Hillyer 的这篇出色的文章，描述两者：management -hierarchical-data-in-mysql

总之：

树存储在如下表中：

CREATE TABLE nested_category (
        category_id INT AUTO_INCREMENT PRIMARY KEY,
        name VARCHAR(20) NOT NULL,
        lft INT NOT NULL,
        rgt INT NOT NULL
);

查找从根到给定节点的路径（此处为“FLASH”）：

SELECT parent.name
FROM nested_category AS node,
        nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND node.name = 'FLASH'
ORDER BY parent.lft;

查找给定节点的所有子节点（此处为“便携式电子设备”）：

SELECT node.name, (COUNT(parent.name) - (sub_tree.depth + 1)) AS depth
FROM nested_category AS node,
        nested_category AS parent,
        nested_category AS sub_parent,
        (
                SELECT node.name, (COUNT(parent.name) - 1) AS depth
                FROM nested_category AS node,
                        nested_category AS parent
                WHERE node.lft BETWEEN parent.lft AND parent.rgt
                        AND node.name = 'PORTABLE ELECTRONICS'
                GROUP BY node.name
                ORDER BY node.lft
        )AS sub_tree
WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND node.lft BETWEEN sub_parent.lft AND sub_parent.rgt
        AND sub_parent.name = sub_tree.name
GROUP BY node.name
HAVING depth <= 1
ORDER BY node.lft;

重命名为您的文件夹表后

  • TABLE nested_category -> TABLE 文件夹
  • 列 category_id -> 列 id
  • 列名 -> 列标题

解决方案是：

CREATE TABLE folders (
        id INT AUTO_INCREMENT PRIMARY KEY,
        title VARCHAR(20) NOT NULL,
        lft INT NOT NULL,
        rgt INT NOT NULL
);

INSERT INTO folders(id, title, lft, rgt) values(1, 'root', 1, 10);
INSERT INTO folders(id, title, lft, rgt) values(2, 'one', 2, 9);
INSERT INTO folders(id, title, lft, rgt) values(3, 'target', 3, 8);
INSERT INTO folders(id, title, lft, rgt) values(4, 'child one', 4, 5);
INSERT INTO folders(id, title, lft, rgt) values(5, 'child two', 6, 7);
INSERT INTO folders(id, title, lft, rgt) values(6, 'root 2', 11, 16);
INSERT INTO folders(id, title, lft, rgt) values(7, 'other child one', 12, 13);
INSERT INTO folders(id, title, lft, rgt) values(8, 'other child two', 14, 15);

目标路径：

SELECT parent.title
FROM folders AS node,
        folders AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND node.title = 'target'
ORDER BY parent.lft;

目标儿童：

SELECT node.title, (COUNT(parent.title) - (sub_tree.depth + 1)) AS depth
    FROM folders AS node,
            folders AS parent,
            folders AS sub_parent,
            (
              SELECT node.title, (COUNT(parent.title) - 1) AS depth
                    FROM folders AS node,
                            folders AS parent
                    WHERE node.lft BETWEEN parent.lft AND parent.rgt
                            AND node.title = 'target'
                    GROUP BY node.title
                    ORDER BY node.lft
            )AS sub_tree
    WHERE node.lft BETWEEN parent.lft AND parent.rgt
            AND node.lft BETWEEN sub_parent.lft AND sub_parent.rgt
            AND sub_parent.title = sub_tree.title
    GROUP BY node.title
    HAVING depth <= 1
    ORDER BY node.lft;

见sqlfiddle

要在单个查询中获取所有数据，union应该这样做。

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