我已经创建了这个数据库。看起来它工作正常,除了我被告知我的表“事件”不是第三范式。我不明白为什么它不是第三种正常形式。我认为这可能是因为城市和邮政编码应该始终相同,但大城市可以有多个邮政编码,而且我没有看到只为城市和他们的邮政编码创建另一个表的重点,相关到事件表。为什么我的SQL表格不是3个正常格式
同样抱歉,如果某些名称或属性使用系统保留的某些名称命名不正确。我不得不把代码翻译成英文,因为我用我的母语写了它:)。谢谢你的帮助。
Create table [article]
(
[id_article] Integer Identity(1,1) NOT NULL,
[id_author] Integer NOT NULL,
[id_category] Integer NOT NULL,
[title] Nvarchar(50) NOT NULL,
[content] Text NOT NULL,
[date] Datetime NOT NULL,
Primary Key ([id_article])
)
go
Create table [author]
(
[id_author] Integer Identity(1,1) NOT NULL,
[name] Nvarchar(25) NOT NULL,
[lastname] Nvarchar(25) NOT NULL,
[email] Nvarchar(50) NOT NULL, UNIQUE ([email]),
[phone] Integer NOT NULL, UNIQUE ([phone]),
[nick] Nvarchar(20) NOT NULL, UNIQUE ([nick]),
[passwd] Nvarchar(50) NOT NULL,
[acc_number] Integer NOT NULL, UNIQUE ([acc_number]),
Primary Key ([id_author])
)
go
Create table [event]
(
[id_event] Integer Identity(1,1) NOT NULL,
[id_author] Integer NOT NULL,
[name] Nvarchar(50) NOT NULL,
[date] Datetime NOT NULL, UNIQUE ([date]),
[city] Nvarchar(50) NOT NULL,
[street] Nvarchar(50) NOT NULL,
[zip] Integer NOT NULL,
[house_number] Integer NOT NULL,
[number_registered] Integer Default 0 NOT NULL Constraint [number_registered] Check (number_registered <= 20),
Primary Key ([id_event])
)
go
Create table [user]
(
[id_user] Integer Identity(1,1) NOT NULL,
[name] Nvarchar(15) NOT NULL,
[lastname] Nvarchar(25) NOT NULL,
[email] Nvarchar(50) NOT NULL, UNIQUE ([email]),
[phone] Integer NOT NULL, UNIQUE ([phone]),
[passwd] Nvarchar(50) NOT NULL,
[nick] Nvarchar(20) NOT NULL, UNIQUE ([nick]),
Primary Key ([id_user])
)
go
Create table [commentary]
(
[id_commentary] Integer Identity(1,1) NOT NULL,
[content] Text NOT NULL,
[id_article] Integer NOT NULL,
[id_author] Integer NULL,
[id_user] Integer NULL,
Primary Key ([id_commentary])
)
go
Create table [category]
(
[id_category] Integer Identity(1,1) NOT NULL,
[name] Nvarchar(30) NOT NULL,
Primary Key ([id_category])
)
go
Create table [registration]
(
[id_user] Integer NOT NULL,
[id_event] Integer NOT NULL,
Primary Key ([id_user],[id_event])
)
go
Alter table [commentary] add foreign key([id_article]) references [article] ([id_article]) on update no action on delete no action
go
Alter table [article] add foreign key([id_author]) references [author] ([id_author]) on update no action on delete no action
go
Alter table [event] add foreign key([id_author]) references [author] ([id_author]) on update no action on delete no action
go
Alter table [commentary] add foreign key([id_author]) references [author] ([id_author]) on update no action on delete no action
go
Alter table [registration] add foreign key([id_event]) references [event] ([id_event]) on update no action on delete no action
go
Alter table [commentary] add foreign key([id_user]) references [user] ([id_user]) on update no action on delete no action
go
Alter table [registration] add foreign key([id_user]) references [user] ([id_user]) on update no action on delete no action
go
Alter table [article] add foreign key([id_category]) references [category] ([id_category]) on update no action on delete no action
go
编辑: 你认为它可以这样工作吗?我制作了另一张名为“位置”的表格,其中包含以前在事件表中的所有地址信息,并创建了id_event PFK。
Create table [event]
(
[id_event] Integer Identity(1,1) NOT NULL,
[id_author] Integer NOT NULL,
[name] Nvarchar(50) NOT NULL,
[datr] Datetime NOT NULL,
[number_registered] Integer Default 0 NOT NULL Constraint [number_registered] Check (number_registered <= 20),
Primary Key ([id_event])
)
go
Create table [location]
(
[city] Char(1) NOT NULL,
[id_event] Integer NOT NULL,
[street] Char(1) NOT NULL,
[house_number] Char(1) NOT NULL,
[zip] Char(1) NOT NULL,
Primary Key ([id_event])
)
go
Alter table [event] add foreign key([id_auhtor]) references [author] ([id_author]) on update no action on delete no action
go
Alter table [location] add foreign key([id_event]) references [event] ([id_event]) on update no action on delete no action
go
我想说审稿人是正确的这不是...毕竟你没有一个不同的邮政编码,城市和街道表。但是...你想要吗?完美的标准化并不总是“正确”的方式。 – Liath
两个事件是否可以在同一地址发生?如果是这样,我会认为分离应该是所有的地址列到一个单独的表中。但是,通常,仅通过查看表定义,就无法确定表是否处于第三范式。无论他们是否取决于实际的*数据*。 –
@Liath:如果它不在3NF中,则必须有传递依赖。它在哪里? –