MySql查询管理用户权限
MySql query to manage user rights
我实际上卡在了一个 MySql 查询上:
我有两个 table:
访问
版本
版本table用于存放软件版本。我有几个软件可以有多个版本。
来自相同 "family"(= 相同软件)的版本具有共同的根值。 lvl 列描述了软件的版本:v1 是 0 级,v3 是 2 级等等
访问权限table描述了允许用户访问的版本。可否开启权限
vstart_id和vend_id用于描述允许用户访问的第一个和最后一个版本
(我们假设 vstart_id 和 vend_id 指的是具有相同根的版本。)
最后一个约束:
vstart_id | vend_id
--------------------
NULL | NULL The user can access ALL versions
3 | NULL The user can access all versions from the version n°3
NULL | 7 The user cannot access version higher than the version n°7
3 | 6 The user can access all versions from 3 (included) to 6 (included)
我真的不知道如何回答以下问题的查询:
用户 n°2 允许访问哪些版本?
这是数据库结构和一些用于测试的数据
CREATE TABLE IF NOT EXISTS `access` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`enabled` tinyint(1) NOT NULL,
`root` int(11) NOT NULL,
`vstart_id` int(11) DEFAULT NULL,
`vend_id` int(11) DEFAULT NULL,
`user_id` int(11) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=8 ;
INSERT INTO `access` (`id`, `enabled`, `root`, `vstart_id`, `vend_id`, `user_id`) VALUES
(1, 1, 1, 2, 3, 1),
(2, 1, 2, 5, 7, 1),
(3, 1, 2, 4, NULL, 2),
(4, 1, 1, NULL, 2, 2),
(5, 1, 2, NULL, 7, 3),
(7, 1, 1, NULL, NULL, 4);
CREATE TABLE IF NOT EXISTS `version` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`root` int(11) NOT NULL,
`lvl` int(11) NOT NULL,
`title` varchar(32) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=10 ;
INSERT INTO `version` (`id`, `root`, `lvl`, `title`) VALUES
(1, 1, 0, 'A - V1'),
(2, 1, 1, 'A - V2'),
(3, 1, 2, 'A - V3'),
(4, 2, 0, 'B - V1'),
(5, 2, 1, 'B - V2'),
(6, 2, 2, 'B - V3'),
(7, 2, 3, 'B - V4'),
(8, 2, 4, 'B - V5'),
(9, 2, 5, 'B - V6');
在此先感谢
假设一个用户每个根最多有一个访问记录,访问中的开始和结束id指的是version.id:
SELECT v.*
FROM `access` a
INNER JOIN version v ON (v.root = a.root)
WHERE a.user_id = 2
AND v.id >= IFNULL(a.vstart_id, 0)
AND (v.id <= a.vend_id OR a.vend_id IS NULL)
根据链接的根记录中的最小和最大 lvl 获取版本信息
SELECT v.*
FROM version v
INNER JOIN(SELECT MIN(v.lvl) AS MinLvl, MAX(v.lvl) as MaxLvl, v.root
FROM `access` a
INNER JOIN version v ON (v.root = a.root)
WHERE a.user_id = 2
GROUP BY root
) t1 ON (t1.root = v.root AND v.lvl >= t1.MinLvl AND v.lvl <= t1.MaxLvl)
http://sqlfiddle.com/#!2/09b3c/10
根据vstart_id和vend_id的lvl获取版本信息(注意vstart_id的lvl必须低于[=29=的lvl ]):
SELECT v.*
FROM version v
INNER JOIN (SELECT a.root, v.lvl as StartLvl, v2.lvl AS EndLvl
FROM access a
LEFT JOIN version v ON (v.id = a.vstart_id AND v.root = a.root)
LEFT JOIN version v2 ON (v2.id = a.vend_id AND v2.root = a.root)
WHERE a.user_id = 2
) t1 ON (t1.root = v.root
AND CASE WHEN StartLvl IS NULL THEN 1=1 ELSE v.lvl >= StartLvl END
AND CASE WHEN EndLvl IS NULL THEN 1=1 ELSE v.lvl <= EndLvl END
)
我实际上卡在了一个 MySql 查询上:
我有两个 table:
访问
版本
版本table用于存放软件版本。我有几个软件可以有多个版本。
来自相同 "family"(= 相同软件)的版本具有共同的根值。 lvl 列描述了软件的版本:v1 是 0 级,v3 是 2 级等等
访问权限table描述了允许用户访问的版本。可否开启权限
vstart_id和vend_id用于描述允许用户访问的第一个和最后一个版本 (我们假设 vstart_id 和 vend_id 指的是具有相同根的版本。)
最后一个约束:
vstart_id | vend_id
--------------------
NULL | NULL The user can access ALL versions
3 | NULL The user can access all versions from the version n°3
NULL | 7 The user cannot access version higher than the version n°7
3 | 6 The user can access all versions from 3 (included) to 6 (included)
我真的不知道如何回答以下问题的查询:
用户 n°2 允许访问哪些版本?
这是数据库结构和一些用于测试的数据
CREATE TABLE IF NOT EXISTS `access` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`enabled` tinyint(1) NOT NULL,
`root` int(11) NOT NULL,
`vstart_id` int(11) DEFAULT NULL,
`vend_id` int(11) DEFAULT NULL,
`user_id` int(11) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=8 ;
INSERT INTO `access` (`id`, `enabled`, `root`, `vstart_id`, `vend_id`, `user_id`) VALUES
(1, 1, 1, 2, 3, 1),
(2, 1, 2, 5, 7, 1),
(3, 1, 2, 4, NULL, 2),
(4, 1, 1, NULL, 2, 2),
(5, 1, 2, NULL, 7, 3),
(7, 1, 1, NULL, NULL, 4);
CREATE TABLE IF NOT EXISTS `version` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`root` int(11) NOT NULL,
`lvl` int(11) NOT NULL,
`title` varchar(32) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=10 ;
INSERT INTO `version` (`id`, `root`, `lvl`, `title`) VALUES
(1, 1, 0, 'A - V1'),
(2, 1, 1, 'A - V2'),
(3, 1, 2, 'A - V3'),
(4, 2, 0, 'B - V1'),
(5, 2, 1, 'B - V2'),
(6, 2, 2, 'B - V3'),
(7, 2, 3, 'B - V4'),
(8, 2, 4, 'B - V5'),
(9, 2, 5, 'B - V6');
在此先感谢
假设一个用户每个根最多有一个访问记录,访问中的开始和结束id指的是version.id:
SELECT v.*
FROM `access` a
INNER JOIN version v ON (v.root = a.root)
WHERE a.user_id = 2
AND v.id >= IFNULL(a.vstart_id, 0)
AND (v.id <= a.vend_id OR a.vend_id IS NULL)
根据链接的根记录中的最小和最大 lvl 获取版本信息
SELECT v.*
FROM version v
INNER JOIN(SELECT MIN(v.lvl) AS MinLvl, MAX(v.lvl) as MaxLvl, v.root
FROM `access` a
INNER JOIN version v ON (v.root = a.root)
WHERE a.user_id = 2
GROUP BY root
) t1 ON (t1.root = v.root AND v.lvl >= t1.MinLvl AND v.lvl <= t1.MaxLvl)
http://sqlfiddle.com/#!2/09b3c/10
根据vstart_id和vend_id的lvl获取版本信息(注意vstart_id的lvl必须低于[=29=的lvl ]):
SELECT v.*
FROM version v
INNER JOIN (SELECT a.root, v.lvl as StartLvl, v2.lvl AS EndLvl
FROM access a
LEFT JOIN version v ON (v.id = a.vstart_id AND v.root = a.root)
LEFT JOIN version v2 ON (v2.id = a.vend_id AND v2.root = a.root)
WHERE a.user_id = 2
) t1 ON (t1.root = v.root
AND CASE WHEN StartLvl IS NULL THEN 1=1 ELSE v.lvl >= StartLvl END
AND CASE WHEN EndLvl IS NULL THEN 1=1 ELSE v.lvl <= EndLvl END
)