Question

我有一个用户项目，放在某种层次结构树中。用户可以拥有多个孩子和多个父母。组织结构图如下所示：

此层次结构使用数据库中的2个表user和position。这里是重现具有少量灯具的表格的SQL，以重现下面提供的组织图像：

-- phpMyAdmin SQL Dump
-- version 4.8.0
-- https://www.phpmyadmin.net/
--
-- Hôte : db
-- Généré le :  jeu. 24 mai 2018 à 07:39
-- Version du serveur :  10.2.14-MariaDB-10.2.14+maria~jessie
-- Version de PHP :  7.2.4

SET SQL_MODE = "NO_AUTO_VALUE_ON_ZERO";
SET AUTOCOMMIT = 0;
START TRANSACTION;
SET time_zone = "+00:00";

CREATE DATABASE IF NOT EXISTS `hierarchy` DEFAULT CHARACTER SET latin1 COLLATE latin1_swedish_ci;
USE `hierarchy`;

DROP TABLE IF EXISTS `position`;
CREATE TABLE `position` (
  `id` int(11) NOT NULL,
  `user_id` int(11) NOT NULL,
  `parent_id` int(11) DEFAULT NULL,
  `uuid` char(36) COLLATE utf8mb4_unicode_ci NOT NULL COMMENT '(DC2Type:guid)',
  `type` varchar(255) COLLATE utf8mb4_unicode_ci NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci;

INSERT INTO `position` (`id`, `user_id`, `parent_id`, `uuid`, `type`) VALUES
(2, 3, NULL, '7db54d6a-6a89-3fe2-87ba-c0dd96d4f664', 'TYPE_HR_PLUS'),
(3, 4, 3, 'adadab97-c0b8-3f89-a245-4edb077a4579', 'TYPE_HR_PLUS'),
(4, 5, 3, '4efec484-8a1f-3c27-a48f-a8c8412e18db', 'TYPE_HR_PLUS'),
(5, 6, 4, '8027b0cb-3e34-357a-8bff-80e4abbf60ef', 'TYPE_HR_PLUS'),
(6, 7, 4, 'e588ec20-9b0e-3ec6-8164-bdfafdf4b440', 'TYPE_HR'),
(7, 8, 5, '61e537d7-5ed6-36b3-8969-1913cdacb42b', 'TYPE_HR'),
(8, 9, 5, '6c241f8f-1a0a-34dc-8042-591352a764ea', 'TYPE_HR_PLUS'),
(9, 9, 3, 'e03998c4-d95a-3410-9727-099f0c83b3d5', 'TYPE_COLLABORATOR'),
(10, 10, 6, '67de2883-a7b0-3260-bb2a-67d3ffa3dc40', 'TYPE_HR'),
(11, 11, 7, '07d37612-01bf-3eec-b249-41c20d756144', 'TYPE_COLLABORATOR'),
(12, 11, 5, '09f7f6f1-59e2-37c0-a2e4-ae9b58ba6c93', 'TYPE_HR'),
(13, 12, 9, '724b8185-b6eb-3d35-b71f-1d076b3020ed', 'TYPE_HR'),
(14, 13, 9, '97d90e5c-7872-3e1e-be65-0f6b5329f54c', 'TYPE_HR_PLUS'),
(15, 14, 10, '3c357b9d-e162-316b-921c-25157d523e70', 'TYPE_COLLABORATOR'),
(16, 14, 3, '542235c6-462c-4922-a629-303430701000', 'TYPE_COLLABORATOR'),
(17, 15, 10, '2d208936-a7e9-32c1-963f-0df7f57ae463', 'TYPE_COLLABORATOR'),
(18, 16, 11, '00b678a2-54cd-3f37-9381-0cd92acea079', 'TYPE_COLLABORATOR'),
(19, 17, 8, 'ec4b8cad-fbce-3692-8d4c-f48f7ffa6452', 'TYPE_COLLABORATOR'),
(20, 18, 12, '48f0c85c-3801-3275-8978-5b2573fd7e0b', 'TYPE_COLLABORATOR'),
(21, 18, 10, 'd6129dec-8823-3304-8dfe-819118073d1f', 'TYPE_COLLABORATOR'),
(22, 19, 13, '1ec9c202-97d7-3311-9119-f6e9f55d058b', 'TYPE_COLLABORATOR'),
(23, 20, 9, '3ab32120-520b-3d41-80f0-56fd3876eecb', 'TYPE_COLLABORATOR');

DROP TABLE IF EXISTS `user`;
CREATE TABLE `user` (
  `id` int(11) NOT NULL,
  `uuid` char(36) COLLATE utf8mb4_unicode_ci NOT NULL COMMENT '(DC2Type:guid)',
  `first_name` varchar(60) COLLATE utf8mb4_unicode_ci NOT NULL,
  `last_name` varchar(60) COLLATE utf8mb4_unicode_ci NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci;

INSERT INTO `user` (`id`, `uuid`, `first_name`, `last_name`) VALUES
(3, 'fb7304f6-5ade-4c06-83e5-36827a4d7904', 'Albus', 'Dumbledore'),
(4, '60af55ca-e9cd-3d8a-bc8a-c9f7ec99baa1', 'Minerva', 'McGonagall'),
(5, 'a422b364-b68f-324b-8836-d38ca62001f5', 'Severus', 'Rogue'),
(6, 'cbdf94a8-f4e3-360b-ba7f-10ea7e59b85d', 'Filius', 'Flitwick'),
(7, '9bce88c9-7b7d-3db3-8780-b6397bb1263d', 'Remus', 'Lupin'),
(8, 'b9d85cfa-3c46-33c1-8750-53be1c281111', 'Pomona', 'Chourave'),
(9, 'c07d10d7-333f-327a-ab32-201209127b7a', 'Rolanda', 'Bibine'),
(10, '8a7b76a1-5f0c-31ff-a93a-ca478598682b', 'Gilderoy', 'Lockhart'),
(11, '7302700a-0bc5-3ed0-84fc-857e975a5771', 'Alastor', 'Maugrey'),
(12, '7b8722f3-1570-3351-864a-ede9ee3a90e6', 'Sybille', 'Trelawney'),
(13, '12a7683a-00da-46a6-b7fd-a6eae5b10e3c', 'Dolores', 'Ombrage'),
(14, 'f054bdeb-a8b8-41e5-be78-b00d1b9377f3', 'Harry', 'Potter'),
(15, 'b1f696ad-cfa3-3830-8685-c45ff4a4ffda', 'Ronald', 'Weasley'),
(16, '8b2a64e9-ed28-363d-a731-3e15e255cd1d', 'Hermione', 'Granger'),
(17, 'f6aa74f9-2a4a-346d-bf75-3f8b02614c19', 'Neville', 'Londubat'),
(18, '8fbbc431-2555-3718-8055-ad8637260732', 'Drago', 'Malefoy'),
(19, '78a58546-8ae1-330d-9426-c23c74b71e78', 'Luna', 'Lovegood'),
(20, '4739a44b-d1b7-3390-9bbb-cbd2e39b4ec0', 'Ginny', 'Weasley');

ALTER TABLE `position`
  ADD PRIMARY KEY (`id`),
  ADD UNIQUE KEY `uuid_idx` (`uuid`),
  ADD KEY `IDX_462CE4F5727ACA70` (`parent_id`),
  ADD KEY `IDX_462CE4F5A76ED395` (`user_id`);

ALTER TABLE `user`
  ADD PRIMARY KEY (`id`),
  ADD UNIQUE KEY `uuid_idx` (`uuid`);

ALTER TABLE `position`
  MODIFY `id` int(11) NOT NULL AUTO_INCREMENT, AUTO_INCREMENT=1024;

ALTER TABLE `user`
  MODIFY `id` int(11) NOT NULL AUTO_INCREMENT, AUTO_INCREMENT=1027;

ALTER TABLE `position`
  ADD CONSTRAINT `FK_462CE4F5727ACA70` FOREIGN KEY (`parent_id`) REFERENCES `user` (`id`),
  ADD CONSTRAINT `FK_462CE4F5A76ED395` FOREIGN KEY (`user_id`) REFERENCES `user` (`id`);


COMMIT;

需要注意的一件重要事情是position表表示用户与其父项之间的关系，它还存储定义关系类型的字符串（实际上它表示此关系中子项的用户类型）

编辑：我没有提到的一个非常重要的事情是这棵树中没有递归，最多可以有5个级别，如组织形象所示：1 HR_PLUS部门主管，其次是最多3个HR_PLUS或HR水平，然后是1个级别的COLLAB

我想要完成的是以下内容：

对于给定的position uuid，我想要获取低于此关系的每个子用户。

例如，Minerva McGonagall和Albus Dumbledore之间的关系是uuid adadab97-c0b8-3f89-a245-4edb077a4579的第3位。鉴于这个uuid，我想获取Minerva McGonagall以下的每个用户：

+---------------------------------------+
| id    | first_name    | last_name     |
+---------------------------------------+
| 6     | Filius        | Flitwick      |
| 7     | Remus         | Lupin         |
| 10    | Gilderoy      | Lockhart      |
| 11    | Alastor       | Maugrey       |
| 14    | Harry         | Potter        |
| 15    | Ronald        | Weasley       |
| 16    | Hermione      | Granger       |
| 18    | Drago         | Malefoy       |
+---------------------------------------+

到目前为止我写的查询如下：

SELECT u2.id, u2.first_name, u2.last_name 
FROM `user` u1 

JOIN position p1 ON p1.user_id = u1.id
JOIN position p2 ON p2.parent_id = p1.user_id
JOIN user u2 ON u2.id = p2.user_id

WHERE p1.uuid = 'adadab97-c0b8-3f89-a245-4edb077a4579'
AND p1.type IN ('TYPE_HR', 'TYPE_HR_PLUS')

我在type属性上添加了一个过滤器，因为如果type不是HR或HR_PLUS，我不希望查询返回任何内容。现在，查询返回Minerva McGonagall（Filius和Remus）正下方的用户。我无法想出如何将其余用户归还给他们。

任何帮助将不胜感激。

我只是使用UNION成功获取预期数据，因为这不是递归树：

SELECT u2.id, u2.first_name, u2.last_name 
FROM `user` u1 

JOIN position p1 ON p1.user_id = u1.id
JOIN position p2 ON p2.parent_id = p1.user_id
JOIN user u2 ON u2.id = p2.user_id

WHERE p1.uuid = 'adadab97-c0b8-3f89-a245-4edb077a4579'
AND p1.type IN ('TYPE_HR', 'TYPE_HR_PLUS')

UNION

SELECT u2.id, u2.first_name, u2.last_name 
FROM `user` u1 

JOIN position p1 ON p1.user_id = u1.id
JOIN position p2 ON p2.parent_id = p1.user_id
JOIN position p3 ON p3.parent_id = p2.user_id
JOIN user u2 ON u2.id = p3.user_id

WHERE p1.uuid = 'adadab97-c0b8-3f89-a245-4edb077a4579'
AND p1.type IN ('TYPE_HR', 'TYPE_HR_PLUS')

UNION

SELECT u2.id, u2.first_name, u2.last_name 
FROM `user` u1 

JOIN position p1 ON p1.user_id = u1.id
JOIN position p2 ON p2.parent_id = p1.user_id
JOIN position p3 ON p3.parent_id = p2.user_id
JOIN position p4 ON p4.parent_id = p3.user_id
JOIN user u2 ON u2.id = p4.user_id

WHERE p1.uuid = 'adadab97-c0b8-3f89-a245-4edb077a4579'
AND p1.type IN ('TYPE_HR', 'TYPE_HR_PLUS')

但这看起来很重，我很确定有一种更简单的方法。

Answer 1

由于您已经找到了联合解决方案，因此这里有一个使用循环（和伪代码）的替代方案：

创建一个表来保存结果，让我们称之为r。

Insert into r the results of your first query, along with a field named level with a value of 1.
set i=2
while (at least a line of r exists where level=i-1) do the following:
insert into r (level=i)
   a modification of your select based on selecting positions from r 
   where level=i-1

Answer 2

这是你正在寻找的吗？ DB Fiddle

with recursive cte as (
  select `user_id` as `ancestor_id`
  /* , 0 as `degreesOfSeperation` */
  , `user_id` 
  from `position`
  union all
  select c.`ancestor_id` 
  /* , c.`degreesOfSeperation` + 1 */
  , p.`user_id` 
  from `cte` as c
  inner join `position` as p
  on p.`parent_id` = c.`user_id`
)
select u.id, u.first_name, u.last_name
from cte c
inner join `user` a on a.`id` = c.`ancestor_id`
inner join `position` p on p.`user_id` = a.`id`
inner join `user` u on u.`id` = c.`user_id`
/* WHERE c.`degreesOfSeperation` > 0 */
WHERE c.`user_id` != c.`ancestor_id` 
AND p.`uuid` = 'adadab97-c0b8-3f89-a245-4edb077a4579'
AND p.`type` IN ('TYPE_HR', 'TYPE_HR_PLUS')
order by c.`degreesOfSeperation`
;

输出

id first_name last_name

Id | First_Name | Last_Name
---+------------+----------  
 6 | Filius     | Flitwick 
 7 | Remus      | Lupin 
10 | Gilderoy   | Lockhart 
11 | Alastor    | Maugrey 
18 | Drago      | Malefoy 
14 | Harry      | Potter 
15 | Ronald     | Weasley 
16 | Hermione   | Granger

大部分内容可能已经为您所熟悉。处理层次结构的逻辑是recursive cte。这有两部分：

第1部分

选择user_id为ancestor_id ，0为degreesOfSeperation ，user_id 来自position

这会从位置表中获取所有用户，将其记录设置为ancestor记录;即我们种植树木的根。我们将degreesOfSeperation设置为0表示这是根记录;即用户和祖先是同一个的那个。

第2部分

  select c.`ancestor_id` 
  , c.`degreesOfSeperation` + 1
  , p.`user_id` 
  from `cte` as c
  inner join `position` as p
  on p.`parent_id` = c.`user_id`

这是递归部分;对于已经添加到树中的每个用户（即从根语句;或者从填充cte表的最后一次迭代），我们现在获取位置父母与之前迭代的用户匹配的任何其他位置。我们在degreesOfSeparation添加一个，这样我们就可以知道这条记录离根祖先有多远。

DegressOfSeparation实际上并不需要（事实上，在上面的陈述中，我已经对此进行了评论）;但我倾向于包含它，因为它对于调试/查看正在进行的操作非常有用。这是cte没有任何过滤器的样子，可以帮助您了解正在发生的事情：

With recursive cte as (
  select `user_id` as `ancestor_id`
  , 0 as `degreesOfSeperation`
  , `user_id` 
  from `position`
  union all
  select c.`ancestor_id` 
  , c.`degreesOfSeperation` + 1
  , p.`user_id` 
  from `cte` as c
  inner join `position` as p
  on p.`parent_id` = c.`user_id`
)
select *
from cte c
order by c.`ancestor_id`
, c.`degreesOfSeperation`
, c.`user_id`
;

Results here

从多个连接层次结构

2 个答案: