使用 1:1 链接表和 CTE（例如，对用户和组的权限）查找与其他关联的 SQL 行

Question

问题

我需要确定用户 U 是否可以根据存储在 MySQL tables（在 MariaDB 10.2 上）的权限使用以下内容访问实体 E规则：

可以将权限授予用户和组。
permissions are only allow: 一个实体和一个user/group之间的关系的存在意味着它（用户或组）被允许访问该实体.
组成员资格：一个用户可以是一个或多个组的一部分。
组继承：一个组可以有一个父组，父组的权限适用于子组。

tables 已经到位，我理解我的问题的方式归结为： “在用户行 U 和实体行 E 之间是否存在 link（通过任何可能的路径）？”

因为 table 没有图表（请参阅下面的详细信息）并且因为组继承我想我可以在这里使用 Common Table 表达式（我不熟悉），递归（对我来说更糟）。

我的尝试

所以我构建了以下 SQL 查询以支持直接实体用户津贴（第一个 CTE：cte_entities_allowed_to_users）

SET @TEST_EID = 302;
SET @TEST_UID = 103;

WITH
cte_entities_allowed_to_users AS (
    SELECT
        DISTINCT
        entities.eid,
        entity_user_access.uid
    FROM
        entity_user_access
        JOIN entities
            ON entity_user_access.eid = entities.eid
)
SELECT
    COUNT(*)
FROM
    cte_entities_allowed_to_users
WHERE
    cte_entities_allowed_to_users.eid = @TEST_EID
    AND cte_entities_allowed_to_users.uid = @TEST_UID

并且还构建了以下其他 CTE：

cte_groups_of_users AS (
    SELECT
        DISTINCT
        users.uid,
        groups.gid
    FROM
        groups
        JOIN group_members
            ON group_members.gid = groups.gid
        JOIN users
            ON users.uid = group_members.uid
)
,
cte_entities_allowed_to_groups AS (
    SELECT
        DISTINCT
        entities.eid,
        entity_group_access.gid
    FROM
        entity_group_access
        JOIN entities
            ON entity_group_access.eid = entities.eid
)

我奋斗的地方

但我不确定：

如何一起使用它们
也不知道如何构建另一个 CTE（主要是使用 group_of_groups table）
CTE 是否是解决此类问题的方法

我所期望的（“真相table”）

对于以下 @TEST_EID 和 @TEST_UID 对，这里是 SELECT（COUNT(*) 字段）的预期返回值：

`@TEST_EID`	`@TEST_UID`	Return	Why
301	101	≥1	`entity_user_access` has a `(301,101)` row
301	102	≥1	`entity_user_access` has a `(301,102)` row
301	103	=0
302	101	=0
302	102	≥1	`entity_group_access` has a `(302,201)` row and `group_members` has a `(201,102)` row
302	103	≥1	`entity_user_access` has a `(302,103)` row
303	101	=0
303	102	≥1	`(303,201)` row in `entity_group_access` and `group_members` has a `(201,102)` row
303	103	≥1	`entity_user_access` has a `(303,103)` row
304	101	≥1	`(304,205)` row in `entity_group_access` and `group_of_groups` has a`(204,205)` row and `group_members` has a `(204,101)` row
304	102	≥1	`(304,201)` row in `entity_group_access` and `group_members` has a `(201,102)` row
304	103	≥1	`(304,205)` row in `entity_group_access` and `group_of_groups` has a`(202,205)` row and `group_members` has a `(202,103)` row

附件 1/1：Table详细信息

关系：


      [users]───────────────[group_members]─────[groups]──┐
         │                       │                 │      │
         │                       │                 └──[group_of_groups]
[entity_user_access]   [entity_group_access]
         │                       │
         └────────┐    ┌─────────┘
                  │    │
                [entities]

一些示例数据集：

users table:

uid	uname
101	Alice
102	Bob
103	Charlie

groups table:

gid	gname
201	Administrators
202	Users
203	Operators
204	Guests
205	X-Mas event

entities table:

eid	ename
301	Foo
302	Bar
303	Qux
304	Snow

entity_user_access table:

eid	uid
301	101
301	102
302	103
303	103

entity_group_access table:

eid	gid
301	201
302	201
303	201
304	201
302	203
304	205

group_members table:

gid	uid	Comment (not part of data)
201	102	Bob is an admin
203	102	Bob is also an operator
202	103	Charlie is an user
204	101	Alice is a guest

group_of_groups table:

gid	parent_gid	Comment (not part of data)
201	203	Admins (201) are Operators (203)
202	205	Users (202) are in the X-Mas event group (205)
204	205	Guests (204) are in the X-Mas event group (205)

完整 SQL:

-- Structure
CREATE TABLE `users` (
    `uid` INT(11) NOT NULL,
    `uname` TINYTEXT NOT NULL,
    PRIMARY KEY (`uid`)
);
CREATE TABLE `groups` (
    `gid` INT(11) NOT NULL,
    `gname` TINYTEXT NOT NULL,
    PRIMARY KEY (`gid`)
);
CREATE TABLE `entities` (
    `eid` INT(11) NOT NULL,
    `ename` TINYTEXT NOT NULL,
    PRIMARY KEY (`eid`)
);
CREATE TABLE `entity_user_access` (
    `eid` INT(11) NOT NULL,
    `uid` INT(11) NOT NULL,
    PRIMARY KEY (`eid`, `uid`),
    CONSTRAINT `FK_eua_entity` FOREIGN KEY (`eid`) REFERENCES `entities` (`eid`),
    CONSTRAINT `FK_eua_user` FOREIGN KEY (`uid`) REFERENCES `users` (`uid`)
);
CREATE TABLE `entity_group_access` (
    `eid` INT(11) NOT NULL,
    `gid` INT(11) NOT NULL,
    PRIMARY KEY (`eid`, `gid`),
    CONSTRAINT `FK_ega_entity` FOREIGN KEY (`eid`) REFERENCES `entities` (`eid`),
    CONSTRAINT `FK_ega_group` FOREIGN KEY (`gid`) REFERENCES `groups` (`gid`)
);
CREATE TABLE `group_members` (
    `gid` INT(11) NOT NULL,
    `uid` INT(11) NOT NULL,
    PRIMARY KEY (`gid`, `uid`),
    CONSTRAINT `FK_gm_group` FOREIGN KEY (`gid`) REFERENCES `groups` (`gid`),
    CONSTRAINT `FK_gm_user` FOREIGN KEY (`uid`) REFERENCES `users` (`uid`)
);
CREATE TABLE `group_of_groups` (
    `gid` INT(11) NOT NULL,
    `parent_gid` INT(11) NOT NULL,
    PRIMARY KEY (`gid`, `parent_gid`),
    CONSTRAINT `FK_gog_group` FOREIGN KEY (`gid`) REFERENCES `groups` (`gid`),
    CONSTRAINT `FK_gog_parent_group` FOREIGN KEY (`parent_gid`) REFERENCES `groups` (`gid`)
);

-- Data
/*!40014 SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0 */;
INSERT INTO `users` VALUES
    (101, 'Alice'),
    (102, 'Bob'),
    (103, 'Charlie');
INSERT INTO `groups` VALUES
    (201, 'Administrators'),
    (202, 'Users'),
    (203, 'Operators'),
    (204, 'Guests'),
    (205, 'X-Mas event');
INSERT INTO `entities` VALUES
    (301, 'Foo'),
    (302, 'Bar'),
    (303, 'Qux'),
    (304, 'Snow');
INSERT INTO `entity_user_access` VALUES
    (301, 101),
    (301, 102),
    (302, 103),
    (303, 103);
INSERT INTO `entity_group_access` VALUES
    (301, 201),
    (302, 201),
    (302, 203),
    (303, 201),
    (304, 201),
    (304, 205);
INSERT INTO `group_members` VALUES
    (201, 102),
    (202, 103),
    (203, 102),
    (204, 101);
INSERT INTO `group_of_groups` VALUES
    (201, 203),
    (202, 205),
    (204, 205);
/*!40014 SET FOREIGN_KEY_CHECKS=IFNULL(@OLD_FOREIGN_KEY_CHECKS, 1) */;

Answer 1

CTE 只是子查询的替代品。

但是你的处理方式有点离谱

您的群组 table 需要与第一个群组具有相同的结构，因此您必须加入其他 table 群组才能获得 eid 和 uid

下一步是 UNION 结果，以便它们垂直连接

在那个工会上你运行你的计数查询

查看示例

SET @TEST_EID = 302;
SET @TEST_UID = 103;

SELECT COUNT(*)
FROM
(    SELECT
        DISTINCT
        entities.eid,
        entity_user_access.uid
    FROM
        entity_user_access
        JOIN entities
            ON entity_user_access.eid = entities.eid
UNION
    SELECT
        DISTINCT
        entities.eid,
        users.uid
    FROM
        `groups`
        JOIN group_members
            ON group_members.gid = groups.gid
        JOIN users
            ON users.uid = group_members.uid
         JOIN 
            entity_group_access
            ON groups.gid = entity_group_access.gid
         JOIN entities
            ON entity_group_access.eid = entities.eid) t1
WHERE eid = @TEST_EID and uid = @TEST_UID

| COUNT(*) |
| -------: |
|        1 |

db<>fiddle here

使用 1:1 链接表和 CTE（例如，对用户和组的权限）查找与其他关联的 SQL 行

Find SQL rows associated to other using 1:1 linking tables and CTE (eg. permissions on users and groups)

mysql

mariadb

common-table-expression

问题

我的尝试

我奋斗的地方

我所期望的（“真相table”）

附件 1/1：Table详细信息