我有这些表格:
Emp_Tutor: Tutor table.
Emp_Tu_De :Tutor details table.
Emp_Tu_St: Tutor status table.
Emp_School: School table.
我们知道学校有很多导师,每个导师可能在一所学校或另一所学校工作,也许有两三所学校。 所以导师表我把它作为学校和导师详细信息之间的破损表格>
导师状态表我们创建它以插入导师教学状态(课程,课程,教学时间,课程等)。
所以我的问题:
我可以在Tutor表中添加主键以建立(Tutor表和Tutor状态表)之间的关系吗?
不要忘记导师桌是一种破碎的关系
look at image attachment.
答案 0 :(得分:0)
我发现尝试存储 status 等内容通常是个错误。例如,就业"状态" of" current","前"," rehire"通常更好地实施为具有开始日期和结束日期的就业表。
损坏的表格和断开的关系在数据库设计中不是正常的英语术语。我不确定你的意思。
PostgreSQL代码如下。 SQL Server将使用datetime数据类型代替标准SQL的时间戳数据类型。可能还有其他一些小差异。
-- Nothing surprising here.
create table schools (
school_id integer primary key,
school_name varchar(20) not null unique
-- other columns go here
);
-- Nothing surprising here.
create table tutors (
tutor_id integer primary key,
tutor_name varchar(20) not null
-- other columns go here
);
-- Nothing surprising here.
create table tutor_details (
tutor_id integer primary key references tutors (tutor_id),
tutor_phone varchar(15)
-- other columns go here
);
-- Predicate: School <school_id> employed <tutor_id>
-- starting on <start_date> and ending on <end_date>.
-- Allows multiple periods of employment.
create table school_tutors (
school_id integer not null references schools (school_id),
tutor_id integer not null references tutors (tutor_id),
start_date date not null default current_date,
end_date date not null default '9999-12-31',
-- You can make a good, practical argument for including end_date
-- in the primary key, but that's a different issue.
primary key (school_id, tutor_id, start_date)
);
-- Only makes sense in the context of employment, so a composite
-- foreign key references school_tutors. In production, I'd try
-- to use more check constraints on the timestamps.
create table tutor_office_hours (
school_id integer not null,
tutor_id integer not null,
start_date date not null,
foreign key (school_id, tutor_id, start_date)
references school_tutors (school_id, tutor_id, start_date),
office_hours_start_time timestamp not null,
office_hours_end_time timestamp not null
check (office_hours_end_time > office_hours_start_time),
primary key (school_id, tutor_id, office_hours_start_time)
);