Question

我有两个表权限和多对多关系组

CREATE TABLE `permissions` (
 `Permission_Id` int(11) NOT NULL AUTO_INCREMENT,
 `Permission_Name` varchar(50) DEFAULT NULL,
 PRIMARY KEY (`Permission_Id`)
)

群组表

CREATE TABLE `groups` (
 `Group_Id` int(11) NOT NULL AUTO_INCREMENT,
 `Group_Desc` varchar(100) DEFAULT NULL,
 PRIMARY KEY (`Group_Id`)
)

我很困惑如何实现多对多的关系

最好在新表中创建Group_id和Permission_id的复合主键

或者创建一个新表＆amp;使用join关键字从两个表中选择列。

Answer 1

来自my blog：

这样做。

CREATE TABLE XtoY (
    # No surrogate id for this table
    x_id MEDIUMINT UNSIGNED NOT NULL,   -- For JOINing to one table
    y_id MEDIUMINT UNSIGNED NOT NULL,   -- For JOINing to the other table
    # Include other fields specific to the 'relation'
    PRIMARY KEY(x_id, y_id),            -- When starting with X
    INDEX      (y_id, x_id)             -- When starting with Y
) ENGINE=InnoDB;

注意：

⚈  Lack of an AUTO_INCREMENT id for this table -- The PK given is the 'natural' PK; there is no good reason for a surrogate.
⚈  "MEDIUMINT" -- This is a reminder that all INTs should be made as small as is safe (smaller ⇒ faster). Of course the declaration here must match the definition in the table being linked to.
⚈  "UNSIGNED" -- Nearly all INTs may as well be declared non-negative
⚈  "NOT NULL" -- Well, that's true, isn't it?
⚈  "InnoDB" -- More effecient than MyISAM because of the way the PRIMARY KEY is clustered with the data in InnoDB.
⚈  "INDEX(y_id, x_id)" -- The PRIMARY KEY makes it efficient to go one direction; this index makes the other direction efficient. No need to say UNIQUE; that would be extra effort on INSERTs.
⚈  In the secondary index, saying just INDEX(y_id) would work because it would implicit include x_id. But I would rather make it more obvious that I am hoping for a 'covering' index.

要有条件地插入新链接，请使用IODKU

请注意，如果此表中有AUTO_INCREMENT，IODKU会“烧掉”＃34; ids很快。

更多

FOREIGN KEY隐含地在所涉及的列上创建索引。

PRIMARY KEY(a,b)（1）表示组合(a,b)为UNIQUE，（2）按(a,b)对数据进行排序。

INDEX(a), INDEX(b)（由FOREIGN KEY生成或手动生成）与INDEX(a,b)相同。

InnoDB真的需要一个PRIMARY KEY，所以你也可以说PRIMARY KEY (a,b) 而不是 UNIQUE(a,b)。

Answer 2

我知道解决方案。

我需要创建＆＃34;交汇点＆＃34;在这种情况下，表可以保持多对多关系。

CREATE TABLE Groups_Permissions
(
Group_Id INT,
Permission_Id INT,
)

Group_Id和Permmission_Id的组合应该是UNIQUE并且具有FK到组和权限表。

设计多对多实体关系的最佳方式

2 个答案: