如何从table中得到所有没有父节点的节点？

Question

如何从 table 中获取所有没有父节点的节点？[=14=]

我正在使用 (Closure Table) 模板。

我需要获取所有没有父节点的节点。

考虑每个节点都引用自身这一事实。也就是说，每个节点本身就是相对于自身的父子节点。

看起来像这样。

下面我展示的是一个测试数据库。

-- phpMyAdmin SQL Dump
-- version 5.1.0
-- https://www.phpmyadmin.net/
--
-- Host: localhost
-- Generation Time: Oct 15, 2021 at 06:30 PM
-- Server version: 8.0.25
-- PHP Version: 8.0.3

SET SQL_MODE = "NO_AUTO_VALUE_ON_ZERO";
START TRANSACTION;
SET time_zone = "+00:00";


/*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */;
/*!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS */;
/*!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION */;
/*!40101 SET NAMES utf8mb4 */;

--
-- Database: `closure`
--

-- --------------------------------------------------------

--
-- Table structure for table `category_name`
--

CREATE TABLE `category_name` (
  `id` bigint NOT NULL,
  `name` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_ru_0900_ai_ci DEFAULT NULL,
  `level` int DEFAULT '0'
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_ru_0900_ai_ci;

--
-- Dumping data for table `category_name`
--

INSERT INTO `category_name` (`id`, `name`, `level`) VALUES
(1, 'Electronics', 0),
(2, 'TV sets', 0),
(3, 'Sensory', 0),
(4, 'Backlit', 0),
(5, 'On wheels', 0),
(6, 'Anti-glare', 0),
(7, 'Super thin', 0),
(8, 'Wall', 0),
(9, 'Telephone', 0),
(10, 'Shell', 0),
(11, 'Button', 0),
(12, 'Sensory', 0),
(13, 'Retractable', 0),
(14, 'Auto', 0),
(15, 'Manual', 0),
(16, 'For home', 0),
(17, 'For bathroom', 0),
(18, 'Rug', 0),
(19, 'With animals', 0),
(20, 'Obstruction', 0),
(21, 'Soap dish ', 0),
(22, 'With holes', 0),
(23, 'Transparent', 0),
(24, 'For kitchen', 0),
(25, 'Pans', 0),
(26, 'With handles', 0),
(27, 'Non-stick', 0),
(28, 'pans', 0),
(29, 'Steam', 0),
(30, 'With lids', 0),
(31, 'Test', 0);

-- --------------------------------------------------------

--
-- Table structure for table `tree_path`
--

CREATE TABLE `tree_path` (
  `children` bigint NOT NULL,
  `parent` bigint NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_ru_0900_ai_ci;

--
-- Dumping data for table `tree_path`
--

INSERT INTO `tree_path` (`children`, `parent`) VALUES
(1, 1),
(2, 1),
(3, 1),
(4, 1),
(5, 1),
(6, 1),
(7, 1),
(8, 1),
(9, 1),
(10, 1),
(11, 1),
(12, 1),
(13, 1),
(14, 1),
(15, 1),
(2, 2),
(3, 2),
(4, 2),
(5, 2),
(6, 2),
(7, 2),
(8, 2),
(3, 3),
(4, 3),
(5, 3),
(4, 4),
(5, 5),
(6, 6),
(7, 6),
(8, 6),
(7, 7),
(8, 8),
(9, 9),
(10, 9),
(11, 9),
(12, 9),
(13, 9),
(14, 9),
(15, 9),
(10, 10),
(11, 10),
(12, 10),
(11, 11),
(12, 12),
(13, 13),
(14, 13),
(15, 13),
(14, 14),
(15, 15),
(16, 16),
(17, 16),
(18, 16),
(19, 16),
(20, 16),
(21, 16),
(22, 16),
(23, 16),
(24, 16),
(25, 16),
(26, 16),
(27, 16),
(28, 16),
(29, 16),
(30, 16),
(17, 17),
(18, 17),
(19, 17),
(20, 17),
(21, 17),
(22, 17),
(23, 17),
(18, 18),
(19, 18),
(20, 18),
(19, 19),
(20, 20),
(21, 21),
(22, 21),
(23, 21),
(22, 22),
(23, 23),
(24, 24),
(25, 24),
(26, 24),
(27, 24),
(28, 24),
(29, 24),
(30, 24),
(25, 25),
(26, 25),
(27, 25),
(26, 26),
(27, 27),
(28, 28),
(29, 28),
(30, 28),
(29, 29),
(30, 30);

--
-- Indexes for dumped tables
--

--
-- Indexes for table `category_name`
--
ALTER TABLE `category_name`
  ADD PRIMARY KEY (`id`);

--
-- Indexes for table `tree_path`
--
ALTER TABLE `tree_path`
  ADD PRIMARY KEY (`children`,`parent`),
  ADD KEY `FK_PARENT` (`parent`);

--
-- AUTO_INCREMENT for dumped tables
--

--
-- AUTO_INCREMENT for table `category_name`
--
ALTER TABLE `category_name`
  MODIFY `id` bigint NOT NULL AUTO_INCREMENT, AUTO_INCREMENT=33;

--
-- Constraints for dumped tables
--

--
-- Constraints for table `tree_path`
--
ALTER TABLE `tree_path`
  ADD CONSTRAINT `FK_CHILDREN` FOREIGN KEY (`children`) REFERENCES `category_name` (`id`) ON DELETE CASCADE,
  ADD CONSTRAINT `FK_PARENT` FOREIGN KEY (`parent`) REFERENCES `category_name` (`id`) ON DELETE CASCADE;
COMMIT;

/*!40101 SET CHARACTER_SET_CLIENT=@OLD_CHARACTER_SET_CLIENT */;
/*!40101 SET CHARACTER_SET_RESULTS=@OLD_CHARACTER_SET_RESULTS */;
/*!40101 SET COLLATION_CONNECTION=@OLD_COLLATION_CONNECTION */;

或许答案会是这样

  +----+--------------+-------+----------+--------+
  | id | name         | level | children | parent |
  +----+--------------+-------+----------+--------+
  |  1 | Electronics  |     0 |      1   |     1  |
  |  16| For home     |     0 |      16  |     16 |
  +----+--------------+-------+----------+--------+

Answer 1

我将使用 NOT EXISTS () 来检查 tree_path table 中是否没有父行（除了自身）。

SELECT
  *
FROM
  category_name   c
WHERE
  NOT EXISTS (
    SELECT *
      FROM tree_path   t
     WHERE t.parent != c.id
       AND t.children = c.id
  )
  AND EXISTS (
    SELECT *
      FROM tree_path   t
     WHERE t.parent = c.id
       AND t.children = c.id
  )

演示：https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=55698aa8b51f99d42db63feda15148a7

（编辑以排除不在 tree_path 中的类别）

如何从table中得到所有没有父节点的节点？

How get from the table all the nodes that have no parent nodes?

mysql

sql

transitive-closure-table